Mixing
Elementary proof for $limlimits_{n toinfty}frac{n!e^n}{n^n} = +infty$
$begingroup$
As seen in this question, the series $sumlimits_{n=1}^{infty} dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)
However, more is true: not only are the terms greater than $1$, they blow up to $+infty$! To see this, one can apply Stirling's approximation and get $dfrac{n!e^n}{n^n} sim sqrt{2pi n}$.
My question is: is there an elementary proof for $lim_{n toinfty}dfrac{n!e^n}{n^n} = +infty$?
The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.
(To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
calculus sequences-and-series limits
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
$endgroup$
add a comment |
$begingroup$
As seen in this question, the series $sumlimits_{n=1}^{infty} dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)
However, more is true: not only are the terms greater than $1$, they blow up to $+infty$! To see this, one can apply Stirling's approximation and get $dfrac{n!e^n}{n^n} sim sqrt{2pi n}$.
My question is: is there an elementary proof for $lim_{n toinfty}dfrac{n!e^n}{n^n} = +infty$?
The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.
(To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
calculus sequences-and-series limits
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
$endgroup$
add a comment |
$begingroup$
As seen in this question, the series $sumlimits_{n=1}^{infty} dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)
However, more is true: not only are the terms greater than $1$, they blow up to $+infty$! To see this, one can apply Stirling's approximation and get $dfrac{n!e^n}{n^n} sim sqrt{2pi n}$.
My question is: is there an elementary proof for $lim_{n toinfty}dfrac{n!e^n}{n^n} = +infty$?
The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.
(To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
calculus sequences-and-series limits
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
$endgroup$
As seen in this question, the series $sumlimits_{n=1}^{infty} dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)
However, more is true: not only are the terms greater than $1$, they blow up to $+infty$! To see this, one can apply Stirling's approximation and get $dfrac{n!e^n}{n^n} sim sqrt{2pi n}$.
My question is: is there an elementary proof for $lim_{n toinfty}dfrac{n!e^n}{n^n} = +infty$?
The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.
(To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
calculus sequences-and-series limits
calculus sequences-and-series limits
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
edited Jan 3 at 9:03
Martin Sleziak
44.7k9117272
44.7k9117272
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
asked Sep 15 '13 at 21:11
Laconian ThinkerLaconian Thinker
397314
397314
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Considering the ratio of consecutive elements of
$$
a_n=frac{n!e^n}{n^n}tag{1}
$$
we get
$$
begin{align}
frac{a_n}{a_{n-1}}
&=frac{(n-1)^{n-1}e}{n^{n-1}}\
&=eleft(1-frac1nright)^{n-1}tag{2}
end{align}
$$
Taking the log of this ratio gives
$$
begin{align}
logleft(frac{a_n}{a_{n-1}}right)
&=1+(n-1)logleft(1-frac1nright)\
&=1-(n-1)left(frac1n+frac1{2n^2}+Oleft(frac1{n^3}right)right)\
&=frac1{2n}+Oleft(frac1{n^2}right)tag{3}
end{align}
$$
Summing $(3)$ yields
$$
log(a_n)=frac12H_n+C+Oleft(frac1nright)tag{4}
$$
Since the harmonic series diverges, $(4)$ shows that $log(a_n)$, and hence $a_n$, grows without bound.
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
One can use integrals. The work is a little long, but quite rewarding. Consider $$I_n=int_0^{pi /2}sin^{n}tdt$$
Then $$I_{2n+1}leqslant I_{2n}leqslant I_{2n-1}$$
On the other hand integrating by parts gives
$${I_{2n + 1}} = frac{{2n}}{{2n + 1}}{I_{2n - 1}}$$
It follows that $$1leqslant frac{{{I_{2n}}}}{{{I_{2n + 1}}}}leqslant frac{{{I_{2n - 1}}}}{{{I_{2n + 1}}}}$$ so $$frac{{{I_{2n}}}}{{{I_{2n + 1}}}}to 1$$
We proceed to evaluate that limit in a different way. Integrating by parts like before gives us $${I_{2n}} = frac{{left( {2n - 1} right)!!}}{{left( {2n} right)!!}}frac{pi }{2}$$ $${I_{2n + 1}} = frac{{left( {2n } right)!!}}{{left( {2n+ 1} right)!!}}$$
Thus, we find $$frac{{{I_{2n + 1}}}}{{{I_{2n}}}} = frac{{left( {2n} right)!{!^2}}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}frac{2}{pi } to 1$$ or $$prodlimits_{k = 1}^infty {frac{{4{k^2}}}{{4{k^2} - 1}}} = frac{pi }{2}$$
This is the celebrated Wallis product for $pi$. Now onto Stirling. Consider the limit $$A = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}$$ Then one must have by $nmapsto 2n$ that $$1 = frac{A}{A} = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}frac{{{{left( {2n} right)}^{2n + 1/2}}}}{{left( {2n} right)!{e^{2n}}}} = sqrt 2 mathop {lim }limits_{n to infty } frac{{{n^n}}}{{{e^n}n!}}frac{{left( {2n} right)!!}}{{left( {2n - 1} right)!!}}$$
We now use Wallis product formula. Squaring, this means that $$1 = 2{left( {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} right)^2}left( {mathop {lim }limits_{n to infty } frac{{2n + 1}}{n}} right)left( {mathop {lim }limits_{n to infty } frac{{left( {2n} right)!!left( {2n} right)!!}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}} right)$$ And we thus get Stirling's $$frac{1}{{sqrt {2pi } }} = { {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} }$$
ADD Denote by $a_n$ our sequence above. Then $${a_n} > {a_{n + 1}}$$ is equivalent to $${left( {1 + frac{1}{n}} right)^{n + 1/2}} > e$$ under rearranging. It remains to prove this.
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
$endgroup$
$begingroup$
Re your addition at the end: You actually need $(1+1/n)^{n+1/2}>e$. This is still true, but a bit harder to prove. I don't see an easy proof other than by a Taylor series argument: When $x>0$, get $e^x<1+x+(x/2)^2/(1-x/2)$ by comparing the tail of the series for $e^x$ with a geometric series, then put $x=1/(n+1/2)$ and work the formulas a bit.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 17:07
$begingroup$
Oops. Will look into it later! Thanks.
$endgroup$
– Pedro Tamaroff♦
Sep 17 '13 at 19:41
1
$begingroup$
I found a nice(?) argument: So long as $0<xlefrac13$, we get $$ln(1+x)>x-frac{x^2}2+frac{x^3}3-frac{x^4}4ge x-frac{x^2}2+frac{x^3}4>frac{x}{1+x/2}$$ (by cutting off the alternating series appropriately). Substitute $x=1/n$, and the desired inequality follows.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 20:06
add a comment |
$begingroup$
Similar to robjohn's explanation, one can approach this using the so-called "trapezoidal rule" for estimating integrals.
Thus, consider the integral $int_1^n log(x) ; dx$. The trapezoidal rule estimate gives, over each interval $[m, m+1]$, the estimate
$$int_m^{m+1} log(x) ; dx ;; = ;; frac1{2}(log(m) + log(m+1)) + e_m$$
where the error term $e_m$ is bounded above by a constant times $m^{-2}$. Summing from $m=1$ to $n-1$, we arrive at
$$xlog(x) - x; |_1^n = int_1^n log(x); dx ;; = ;; frac1{2}log(n)+ sum_{1}^{n-1} log(n) + E_n$$
where the error term $E_n = sum_{m=1}^{n-1} e_m$ is some constant $E = sum_{m=1}^infty e_m$ minus an amount that is bounded above by some constant $C$ times $n^{-1}$. We may exponentiate to get $n^n e^{-n}e^{-1} = e^{E_n}(frac{n!}{sqrt{n}})$, which may be massaged into the asymptotic formula
$$frac{e^n n!}{n^n} = (1 + O(n^{-1}))e^{1-E}sqrt{n}$$
where $E$ could be determined if we want, but this gets the job done. See Terry Tao's post here.
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f494776%2felementary-proof-for-lim-limits-n-to-infty-fracnennn-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Considering the ratio of consecutive elements of
$$
a_n=frac{n!e^n}{n^n}tag{1}
$$
we get
$$
begin{align}
frac{a_n}{a_{n-1}}
&=frac{(n-1)^{n-1}e}{n^{n-1}}\
&=eleft(1-frac1nright)^{n-1}tag{2}
end{align}
$$
Taking the log of this ratio gives
$$
begin{align}
logleft(frac{a_n}{a_{n-1}}right)
&=1+(n-1)logleft(1-frac1nright)\
&=1-(n-1)left(frac1n+frac1{2n^2}+Oleft(frac1{n^3}right)right)\
&=frac1{2n}+Oleft(frac1{n^2}right)tag{3}
end{align}
$$
Summing $(3)$ yields
$$
log(a_n)=frac12H_n+C+Oleft(frac1nright)tag{4}
$$
Since the harmonic series diverges, $(4)$ shows that $log(a_n)$, and hence $a_n$, grows without bound.
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
Considering the ratio of consecutive elements of
$$
a_n=frac{n!e^n}{n^n}tag{1}
$$
we get
$$
begin{align}
frac{a_n}{a_{n-1}}
&=frac{(n-1)^{n-1}e}{n^{n-1}}\
&=eleft(1-frac1nright)^{n-1}tag{2}
end{align}
$$
Taking the log of this ratio gives
$$
begin{align}
logleft(frac{a_n}{a_{n-1}}right)
&=1+(n-1)logleft(1-frac1nright)\
&=1-(n-1)left(frac1n+frac1{2n^2}+Oleft(frac1{n^3}right)right)\
&=frac1{2n}+Oleft(frac1{n^2}right)tag{3}
end{align}
$$
Summing $(3)$ yields
$$
log(a_n)=frac12H_n+C+Oleft(frac1nright)tag{4}
$$
Since the harmonic series diverges, $(4)$ shows that $log(a_n)$, and hence $a_n$, grows without bound.
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
Considering the ratio of consecutive elements of
$$
a_n=frac{n!e^n}{n^n}tag{1}
$$
we get
$$
begin{align}
frac{a_n}{a_{n-1}}
&=frac{(n-1)^{n-1}e}{n^{n-1}}\
&=eleft(1-frac1nright)^{n-1}tag{2}
end{align}
$$
Taking the log of this ratio gives
$$
begin{align}
logleft(frac{a_n}{a_{n-1}}right)
&=1+(n-1)logleft(1-frac1nright)\
&=1-(n-1)left(frac1n+frac1{2n^2}+Oleft(frac1{n^3}right)right)\
&=frac1{2n}+Oleft(frac1{n^2}right)tag{3}
end{align}
$$
Summing $(3)$ yields
$$
log(a_n)=frac12H_n+C+Oleft(frac1nright)tag{4}
$$
Since the harmonic series diverges, $(4)$ shows that $log(a_n)$, and hence $a_n$, grows without bound.
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
$endgroup$
Considering the ratio of consecutive elements of
$$
a_n=frac{n!e^n}{n^n}tag{1}
$$
we get
$$
begin{align}
frac{a_n}{a_{n-1}}
&=frac{(n-1)^{n-1}e}{n^{n-1}}\
&=eleft(1-frac1nright)^{n-1}tag{2}
end{align}
$$
Taking the log of this ratio gives
$$
begin{align}
logleft(frac{a_n}{a_{n-1}}right)
&=1+(n-1)logleft(1-frac1nright)\
&=1-(n-1)left(frac1n+frac1{2n^2}+Oleft(frac1{n^3}right)right)\
&=frac1{2n}+Oleft(frac1{n^2}right)tag{3}
end{align}
$$
Summing $(3)$ yields
$$
log(a_n)=frac12H_n+C+Oleft(frac1nright)tag{4}
$$
Since the harmonic series diverges, $(4)$ shows that $log(a_n)$, and hence $a_n$, grows without bound.
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
answered Sep 15 '13 at 22:05
robjohn♦robjohn
266k27304626
266k27304626
add a comment |
add a comment |
$begingroup$
One can use integrals. The work is a little long, but quite rewarding. Consider $$I_n=int_0^{pi /2}sin^{n}tdt$$
Then $$I_{2n+1}leqslant I_{2n}leqslant I_{2n-1}$$
On the other hand integrating by parts gives
$${I_{2n + 1}} = frac{{2n}}{{2n + 1}}{I_{2n - 1}}$$
It follows that $$1leqslant frac{{{I_{2n}}}}{{{I_{2n + 1}}}}leqslant frac{{{I_{2n - 1}}}}{{{I_{2n + 1}}}}$$ so $$frac{{{I_{2n}}}}{{{I_{2n + 1}}}}to 1$$
We proceed to evaluate that limit in a different way. Integrating by parts like before gives us $${I_{2n}} = frac{{left( {2n - 1} right)!!}}{{left( {2n} right)!!}}frac{pi }{2}$$ $${I_{2n + 1}} = frac{{left( {2n } right)!!}}{{left( {2n+ 1} right)!!}}$$
Thus, we find $$frac{{{I_{2n + 1}}}}{{{I_{2n}}}} = frac{{left( {2n} right)!{!^2}}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}frac{2}{pi } to 1$$ or $$prodlimits_{k = 1}^infty {frac{{4{k^2}}}{{4{k^2} - 1}}} = frac{pi }{2}$$
This is the celebrated Wallis product for $pi$. Now onto Stirling. Consider the limit $$A = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}$$ Then one must have by $nmapsto 2n$ that $$1 = frac{A}{A} = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}frac{{{{left( {2n} right)}^{2n + 1/2}}}}{{left( {2n} right)!{e^{2n}}}} = sqrt 2 mathop {lim }limits_{n to infty } frac{{{n^n}}}{{{e^n}n!}}frac{{left( {2n} right)!!}}{{left( {2n - 1} right)!!}}$$
We now use Wallis product formula. Squaring, this means that $$1 = 2{left( {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} right)^2}left( {mathop {lim }limits_{n to infty } frac{{2n + 1}}{n}} right)left( {mathop {lim }limits_{n to infty } frac{{left( {2n} right)!!left( {2n} right)!!}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}} right)$$ And we thus get Stirling's $$frac{1}{{sqrt {2pi } }} = { {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} }$$
ADD Denote by $a_n$ our sequence above. Then $${a_n} > {a_{n + 1}}$$ is equivalent to $${left( {1 + frac{1}{n}} right)^{n + 1/2}} > e$$ under rearranging. It remains to prove this.
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
$endgroup$
$begingroup$
Re your addition at the end: You actually need $(1+1/n)^{n+1/2}>e$. This is still true, but a bit harder to prove. I don't see an easy proof other than by a Taylor series argument: When $x>0$, get $e^x<1+x+(x/2)^2/(1-x/2)$ by comparing the tail of the series for $e^x$ with a geometric series, then put $x=1/(n+1/2)$ and work the formulas a bit.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 17:07
$begingroup$
Oops. Will look into it later! Thanks.
$endgroup$
– Pedro Tamaroff♦
Sep 17 '13 at 19:41
1
$begingroup$
I found a nice(?) argument: So long as $0<xlefrac13$, we get $$ln(1+x)>x-frac{x^2}2+frac{x^3}3-frac{x^4}4ge x-frac{x^2}2+frac{x^3}4>frac{x}{1+x/2}$$ (by cutting off the alternating series appropriately). Substitute $x=1/n$, and the desired inequality follows.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 20:06
add a comment |
$begingroup$
One can use integrals. The work is a little long, but quite rewarding. Consider $$I_n=int_0^{pi /2}sin^{n}tdt$$
Then $$I_{2n+1}leqslant I_{2n}leqslant I_{2n-1}$$
On the other hand integrating by parts gives
$${I_{2n + 1}} = frac{{2n}}{{2n + 1}}{I_{2n - 1}}$$
It follows that $$1leqslant frac{{{I_{2n}}}}{{{I_{2n + 1}}}}leqslant frac{{{I_{2n - 1}}}}{{{I_{2n + 1}}}}$$ so $$frac{{{I_{2n}}}}{{{I_{2n + 1}}}}to 1$$
We proceed to evaluate that limit in a different way. Integrating by parts like before gives us $${I_{2n}} = frac{{left( {2n - 1} right)!!}}{{left( {2n} right)!!}}frac{pi }{2}$$ $${I_{2n + 1}} = frac{{left( {2n } right)!!}}{{left( {2n+ 1} right)!!}}$$
Thus, we find $$frac{{{I_{2n + 1}}}}{{{I_{2n}}}} = frac{{left( {2n} right)!{!^2}}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}frac{2}{pi } to 1$$ or $$prodlimits_{k = 1}^infty {frac{{4{k^2}}}{{4{k^2} - 1}}} = frac{pi }{2}$$
This is the celebrated Wallis product for $pi$. Now onto Stirling. Consider the limit $$A = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}$$ Then one must have by $nmapsto 2n$ that $$1 = frac{A}{A} = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}frac{{{{left( {2n} right)}^{2n + 1/2}}}}{{left( {2n} right)!{e^{2n}}}} = sqrt 2 mathop {lim }limits_{n to infty } frac{{{n^n}}}{{{e^n}n!}}frac{{left( {2n} right)!!}}{{left( {2n - 1} right)!!}}$$
We now use Wallis product formula. Squaring, this means that $$1 = 2{left( {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} right)^2}left( {mathop {lim }limits_{n to infty } frac{{2n + 1}}{n}} right)left( {mathop {lim }limits_{n to infty } frac{{left( {2n} right)!!left( {2n} right)!!}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}} right)$$ And we thus get Stirling's $$frac{1}{{sqrt {2pi } }} = { {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} }$$
ADD Denote by $a_n$ our sequence above. Then $${a_n} > {a_{n + 1}}$$ is equivalent to $${left( {1 + frac{1}{n}} right)^{n + 1/2}} > e$$ under rearranging. It remains to prove this.
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
$endgroup$
$begingroup$
Re your addition at the end: You actually need $(1+1/n)^{n+1/2}>e$. This is still true, but a bit harder to prove. I don't see an easy proof other than by a Taylor series argument: When $x>0$, get $e^x<1+x+(x/2)^2/(1-x/2)$ by comparing the tail of the series for $e^x$ with a geometric series, then put $x=1/(n+1/2)$ and work the formulas a bit.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 17:07
$begingroup$
Oops. Will look into it later! Thanks.
$endgroup$
– Pedro Tamaroff♦
Sep 17 '13 at 19:41
1
$begingroup$
I found a nice(?) argument: So long as $0<xlefrac13$, we get $$ln(1+x)>x-frac{x^2}2+frac{x^3}3-frac{x^4}4ge x-frac{x^2}2+frac{x^3}4>frac{x}{1+x/2}$$ (by cutting off the alternating series appropriately). Substitute $x=1/n$, and the desired inequality follows.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 20:06
add a comment |
$begingroup$
One can use integrals. The work is a little long, but quite rewarding. Consider $$I_n=int_0^{pi /2}sin^{n}tdt$$
Then $$I_{2n+1}leqslant I_{2n}leqslant I_{2n-1}$$
On the other hand integrating by parts gives
$${I_{2n + 1}} = frac{{2n}}{{2n + 1}}{I_{2n - 1}}$$
It follows that $$1leqslant frac{{{I_{2n}}}}{{{I_{2n + 1}}}}leqslant frac{{{I_{2n - 1}}}}{{{I_{2n + 1}}}}$$ so $$frac{{{I_{2n}}}}{{{I_{2n + 1}}}}to 1$$
We proceed to evaluate that limit in a different way. Integrating by parts like before gives us $${I_{2n}} = frac{{left( {2n - 1} right)!!}}{{left( {2n} right)!!}}frac{pi }{2}$$ $${I_{2n + 1}} = frac{{left( {2n } right)!!}}{{left( {2n+ 1} right)!!}}$$
Thus, we find $$frac{{{I_{2n + 1}}}}{{{I_{2n}}}} = frac{{left( {2n} right)!{!^2}}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}frac{2}{pi } to 1$$ or $$prodlimits_{k = 1}^infty {frac{{4{k^2}}}{{4{k^2} - 1}}} = frac{pi }{2}$$
This is the celebrated Wallis product for $pi$. Now onto Stirling. Consider the limit $$A = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}$$ Then one must have by $nmapsto 2n$ that $$1 = frac{A}{A} = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}frac{{{{left( {2n} right)}^{2n + 1/2}}}}{{left( {2n} right)!{e^{2n}}}} = sqrt 2 mathop {lim }limits_{n to infty } frac{{{n^n}}}{{{e^n}n!}}frac{{left( {2n} right)!!}}{{left( {2n - 1} right)!!}}$$
We now use Wallis product formula. Squaring, this means that $$1 = 2{left( {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} right)^2}left( {mathop {lim }limits_{n to infty } frac{{2n + 1}}{n}} right)left( {mathop {lim }limits_{n to infty } frac{{left( {2n} right)!!left( {2n} right)!!}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}} right)$$ And we thus get Stirling's $$frac{1}{{sqrt {2pi } }} = { {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} }$$
ADD Denote by $a_n$ our sequence above. Then $${a_n} > {a_{n + 1}}$$ is equivalent to $${left( {1 + frac{1}{n}} right)^{n + 1/2}} > e$$ under rearranging. It remains to prove this.
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
$endgroup$
One can use integrals. The work is a little long, but quite rewarding. Consider $$I_n=int_0^{pi /2}sin^{n}tdt$$
Then $$I_{2n+1}leqslant I_{2n}leqslant I_{2n-1}$$
On the other hand integrating by parts gives
$${I_{2n + 1}} = frac{{2n}}{{2n + 1}}{I_{2n - 1}}$$
It follows that $$1leqslant frac{{{I_{2n}}}}{{{I_{2n + 1}}}}leqslant frac{{{I_{2n - 1}}}}{{{I_{2n + 1}}}}$$ so $$frac{{{I_{2n}}}}{{{I_{2n + 1}}}}to 1$$
We proceed to evaluate that limit in a different way. Integrating by parts like before gives us $${I_{2n}} = frac{{left( {2n - 1} right)!!}}{{left( {2n} right)!!}}frac{pi }{2}$$ $${I_{2n + 1}} = frac{{left( {2n } right)!!}}{{left( {2n+ 1} right)!!}}$$
Thus, we find $$frac{{{I_{2n + 1}}}}{{{I_{2n}}}} = frac{{left( {2n} right)!{!^2}}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}frac{2}{pi } to 1$$ or $$prodlimits_{k = 1}^infty {frac{{4{k^2}}}{{4{k^2} - 1}}} = frac{pi }{2}$$
This is the celebrated Wallis product for $pi$. Now onto Stirling. Consider the limit $$A = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}$$ Then one must have by $nmapsto 2n$ that $$1 = frac{A}{A} = mathop {lim }limits_{n to infty } frac{{n!{e^n}}}{{{n^{n + 1/2}}}}frac{{{{left( {2n} right)}^{2n + 1/2}}}}{{left( {2n} right)!{e^{2n}}}} = sqrt 2 mathop {lim }limits_{n to infty } frac{{{n^n}}}{{{e^n}n!}}frac{{left( {2n} right)!!}}{{left( {2n - 1} right)!!}}$$
We now use Wallis product formula. Squaring, this means that $$1 = 2{left( {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} right)^2}left( {mathop {lim }limits_{n to infty } frac{{2n + 1}}{n}} right)left( {mathop {lim }limits_{n to infty } frac{{left( {2n} right)!!left( {2n} right)!!}}{{left( {2n - 1} right)!!left( {2n + 1} right)!!}}} right)$$ And we thus get Stirling's $$frac{1}{{sqrt {2pi } }} = { {mathop {lim }limits_{n to infty } frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} }$$
ADD Denote by $a_n$ our sequence above. Then $${a_n} > {a_{n + 1}}$$ is equivalent to $${left( {1 + frac{1}{n}} right)^{n + 1/2}} > e$$ under rearranging. It remains to prove this.
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
edited Sep 29 '13 at 3:02
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
answered Sep 15 '13 at 21:55
Pedro Tamaroff♦Pedro Tamaroff
96.4k10152296
96.4k10152296
$begingroup$
Re your addition at the end: You actually need $(1+1/n)^{n+1/2}>e$. This is still true, but a bit harder to prove. I don't see an easy proof other than by a Taylor series argument: When $x>0$, get $e^x<1+x+(x/2)^2/(1-x/2)$ by comparing the tail of the series for $e^x$ with a geometric series, then put $x=1/(n+1/2)$ and work the formulas a bit.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 17:07
$begingroup$
Oops. Will look into it later! Thanks.
$endgroup$
– Pedro Tamaroff♦
Sep 17 '13 at 19:41
1
$begingroup$
I found a nice(?) argument: So long as $0<xlefrac13$, we get $$ln(1+x)>x-frac{x^2}2+frac{x^3}3-frac{x^4}4ge x-frac{x^2}2+frac{x^3}4>frac{x}{1+x/2}$$ (by cutting off the alternating series appropriately). Substitute $x=1/n$, and the desired inequality follows.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 20:06
add a comment |
$begingroup$
Re your addition at the end: You actually need $(1+1/n)^{n+1/2}>e$. This is still true, but a bit harder to prove. I don't see an easy proof other than by a Taylor series argument: When $x>0$, get $e^x<1+x+(x/2)^2/(1-x/2)$ by comparing the tail of the series for $e^x$ with a geometric series, then put $x=1/(n+1/2)$ and work the formulas a bit.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 17:07
$begingroup$
Oops. Will look into it later! Thanks.
$endgroup$
– Pedro Tamaroff♦
Sep 17 '13 at 19:41
1
$begingroup$
I found a nice(?) argument: So long as $0<xlefrac13$, we get $$ln(1+x)>x-frac{x^2}2+frac{x^3}3-frac{x^4}4ge x-frac{x^2}2+frac{x^3}4>frac{x}{1+x/2}$$ (by cutting off the alternating series appropriately). Substitute $x=1/n$, and the desired inequality follows.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 20:06
$begingroup$
Re your addition at the end: You actually need $(1+1/n)^{n+1/2}>e$. This is still true, but a bit harder to prove. I don't see an easy proof other than by a Taylor series argument: When $x>0$, get $e^x<1+x+(x/2)^2/(1-x/2)$ by comparing the tail of the series for $e^x$ with a geometric series, then put $x=1/(n+1/2)$ and work the formulas a bit.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 17:07
$begingroup$
Re your addition at the end: You actually need $(1+1/n)^{n+1/2}>e$. This is still true, but a bit harder to prove. I don't see an easy proof other than by a Taylor series argument: When $x>0$, get $e^x<1+x+(x/2)^2/(1-x/2)$ by comparing the tail of the series for $e^x$ with a geometric series, then put $x=1/(n+1/2)$ and work the formulas a bit.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 17:07
$begingroup$
Oops. Will look into it later! Thanks.
$endgroup$
– Pedro Tamaroff♦
Sep 17 '13 at 19:41
$begingroup$
Oops. Will look into it later! Thanks.
$endgroup$
– Pedro Tamaroff♦
Sep 17 '13 at 19:41
1
1
$begingroup$
I found a nice(?) argument: So long as $0<xlefrac13$, we get $$ln(1+x)>x-frac{x^2}2+frac{x^3}3-frac{x^4}4ge x-frac{x^2}2+frac{x^3}4>frac{x}{1+x/2}$$ (by cutting off the alternating series appropriately). Substitute $x=1/n$, and the desired inequality follows.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 20:06
$begingroup$
I found a nice(?) argument: So long as $0<xlefrac13$, we get $$ln(1+x)>x-frac{x^2}2+frac{x^3}3-frac{x^4}4ge x-frac{x^2}2+frac{x^3}4>frac{x}{1+x/2}$$ (by cutting off the alternating series appropriately). Substitute $x=1/n$, and the desired inequality follows.
$endgroup$
– Harald Hanche-Olsen
Sep 17 '13 at 20:06
add a comment |
$begingroup$
Similar to robjohn's explanation, one can approach this using the so-called "trapezoidal rule" for estimating integrals.
Thus, consider the integral $int_1^n log(x) ; dx$. The trapezoidal rule estimate gives, over each interval $[m, m+1]$, the estimate
$$int_m^{m+1} log(x) ; dx ;; = ;; frac1{2}(log(m) + log(m+1)) + e_m$$
where the error term $e_m$ is bounded above by a constant times $m^{-2}$. Summing from $m=1$ to $n-1$, we arrive at
$$xlog(x) - x; |_1^n = int_1^n log(x); dx ;; = ;; frac1{2}log(n)+ sum_{1}^{n-1} log(n) + E_n$$
where the error term $E_n = sum_{m=1}^{n-1} e_m$ is some constant $E = sum_{m=1}^infty e_m$ minus an amount that is bounded above by some constant $C$ times $n^{-1}$. We may exponentiate to get $n^n e^{-n}e^{-1} = e^{E_n}(frac{n!}{sqrt{n}})$, which may be massaged into the asymptotic formula
$$frac{e^n n!}{n^n} = (1 + O(n^{-1}))e^{1-E}sqrt{n}$$
where $E$ could be determined if we want, but this gets the job done. See Terry Tao's post here.
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
$endgroup$
add a comment |
$begingroup$
Similar to robjohn's explanation, one can approach this using the so-called "trapezoidal rule" for estimating integrals.
Thus, consider the integral $int_1^n log(x) ; dx$. The trapezoidal rule estimate gives, over each interval $[m, m+1]$, the estimate
$$int_m^{m+1} log(x) ; dx ;; = ;; frac1{2}(log(m) + log(m+1)) + e_m$$
where the error term $e_m$ is bounded above by a constant times $m^{-2}$. Summing from $m=1$ to $n-1$, we arrive at
$$xlog(x) - x; |_1^n = int_1^n log(x); dx ;; = ;; frac1{2}log(n)+ sum_{1}^{n-1} log(n) + E_n$$
where the error term $E_n = sum_{m=1}^{n-1} e_m$ is some constant $E = sum_{m=1}^infty e_m$ minus an amount that is bounded above by some constant $C$ times $n^{-1}$. We may exponentiate to get $n^n e^{-n}e^{-1} = e^{E_n}(frac{n!}{sqrt{n}})$, which may be massaged into the asymptotic formula
$$frac{e^n n!}{n^n} = (1 + O(n^{-1}))e^{1-E}sqrt{n}$$
where $E$ could be determined if we want, but this gets the job done. See Terry Tao's post here.
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
$endgroup$
add a comment |
$begingroup$
Similar to robjohn's explanation, one can approach this using the so-called "trapezoidal rule" for estimating integrals.
Thus, consider the integral $int_1^n log(x) ; dx$. The trapezoidal rule estimate gives, over each interval $[m, m+1]$, the estimate
$$int_m^{m+1} log(x) ; dx ;; = ;; frac1{2}(log(m) + log(m+1)) + e_m$$
where the error term $e_m$ is bounded above by a constant times $m^{-2}$. Summing from $m=1$ to $n-1$, we arrive at
$$xlog(x) - x; |_1^n = int_1^n log(x); dx ;; = ;; frac1{2}log(n)+ sum_{1}^{n-1} log(n) + E_n$$
where the error term $E_n = sum_{m=1}^{n-1} e_m$ is some constant $E = sum_{m=1}^infty e_m$ minus an amount that is bounded above by some constant $C$ times $n^{-1}$. We may exponentiate to get $n^n e^{-n}e^{-1} = e^{E_n}(frac{n!}{sqrt{n}})$, which may be massaged into the asymptotic formula
$$frac{e^n n!}{n^n} = (1 + O(n^{-1}))e^{1-E}sqrt{n}$$
where $E$ could be determined if we want, but this gets the job done. See Terry Tao's post here.
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
$endgroup$
Similar to robjohn's explanation, one can approach this using the so-called "trapezoidal rule" for estimating integrals.
Thus, consider the integral $int_1^n log(x) ; dx$. The trapezoidal rule estimate gives, over each interval $[m, m+1]$, the estimate
$$int_m^{m+1} log(x) ; dx ;; = ;; frac1{2}(log(m) + log(m+1)) + e_m$$
where the error term $e_m$ is bounded above by a constant times $m^{-2}$. Summing from $m=1$ to $n-1$, we arrive at
$$xlog(x) - x; |_1^n = int_1^n log(x); dx ;; = ;; frac1{2}log(n)+ sum_{1}^{n-1} log(n) + E_n$$
where the error term $E_n = sum_{m=1}^{n-1} e_m$ is some constant $E = sum_{m=1}^infty e_m$ minus an amount that is bounded above by some constant $C$ times $n^{-1}$. We may exponentiate to get $n^n e^{-n}e^{-1} = e^{E_n}(frac{n!}{sqrt{n}})$, which may be massaged into the asymptotic formula
$$frac{e^n n!}{n^n} = (1 + O(n^{-1}))e^{1-E}sqrt{n}$$
where $E$ could be determined if we want, but this gets the job done. See Terry Tao's post here.
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
answered Sep 15 '13 at 22:49
user43208user43208
5,9681832
5,9681832
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f494776%2felementary-proof-for-lim-limits-n-to-infty-fracnennn-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
