Mixing
Proof verification of $left(1+{1over n}right)^n > e^{1-1/n}$ for all $ninBbb N$.
$begingroup$
Prove that for all $ninBbb N$:
$$
left(1+{1over n}right)^n > e^{1-{1over n}}
$$
Suppose the inequality holds.
Using the fact that:
$$
lim_{ntoinfty}left(1+{xover n}right)^n = e^x
$$
and:
$$
left(1+{xover n}right)^n le left(1+{xover n+1}right)^{n+1}
$$
put $x = {1-{1over n}}$ in the above:
$$
e^{1-{1over n}} > left(1+frac{1-{1over n}}{n}right)^n = \
= left(1+frac{n-1}{n^2}right)^n = y_n
$$
Let $x_n$:
$$
x_n = left(1+{1over n}right)^n
$$
Consider the fraction:
$$
frac{x_n}{y_n} = frac{left(1+{1over n}right)^n}{left(1+frac{n-1}{n^2}right)^n} \
=left(frac{n^2(n+1)}{n(n^2 + n - 1)}right)^n \
= left(frac{n^2 + n}{n^2 + n - 1}right)^n > 1
$$
Therefore $x_n > y_n$. But at the same time:
$$
e^{1-{1over n}} > y_n
$$
Thus:
$$
x_n > e^{1-{1over n}} > y_n
$$
Define:
$$
z_n = e^{1-{1over n}}
$$
Now if we take the limit of both sides then by squeeze theorem we get:
$$
lim_{ntoinfty}x_n gelim_{ntoinfty} z_n ge lim_{ntoinfty}y_n
$$
All three limits are equal to $e$ which shows that $z_n$ is squeezed between $x_n$ and $y_n$ meaning that $x_n > z_n$.
I'm not pretty sure the reasoning above is valid, so i'm kindly asking to verify it or point to the flaws. In case the above is wrong could you please show the way to prove the inequality? Thank you!
calculus sequences-and-series limits proof-verification inequality
asked Jan 10 at 16:39
romanroman
2,18521224
$endgroup$
add a comment |
$begingroup$
Prove that for all $ninBbb N$:
$$
left(1+{1over n}right)^n > e^{1-{1over n}}
$$
Suppose the inequality holds.
Using the fact that:
$$
lim_{ntoinfty}left(1+{xover n}right)^n = e^x
$$
and:
$$
left(1+{xover n}right)^n le left(1+{xover n+1}right)^{n+1}
$$
put $x = {1-{1over n}}$ in the above:
$$
e^{1-{1over n}} > left(1+frac{1-{1over n}}{n}right)^n = \
= left(1+frac{n-1}{n^2}right)^n = y_n
$$
Let $x_n$:
$$
x_n = left(1+{1over n}right)^n
$$
Consider the fraction:
$$
frac{x_n}{y_n} = frac{left(1+{1over n}right)^n}{left(1+frac{n-1}{n^2}right)^n} \
=left(frac{n^2(n+1)}{n(n^2 + n - 1)}right)^n \
= left(frac{n^2 + n}{n^2 + n - 1}right)^n > 1
$$
Therefore $x_n > y_n$. But at the same time:
$$
e^{1-{1over n}} > y_n
$$
Thus:
$$
x_n > e^{1-{1over n}} > y_n
$$
Define:
$$
z_n = e^{1-{1over n}}
$$
Now if we take the limit of both sides then by squeeze theorem we get:
$$
lim_{ntoinfty}x_n gelim_{ntoinfty} z_n ge lim_{ntoinfty}y_n
$$
All three limits are equal to $e$ which shows that $z_n$ is squeezed between $x_n$ and $y_n$ meaning that $x_n > z_n$.
I'm not pretty sure the reasoning above is valid, so i'm kindly asking to verify it or point to the flaws. In case the above is wrong could you please show the way to prove the inequality? Thank you!
calculus sequences-and-series limits proof-verification inequality
asked Jan 10 at 16:39
romanroman
2,18521224
$endgroup$
$begingroup$
Your proof appears to be backwards: the very first thing that you do is explicitly assume that your conclusion is true. I've actually no idea where the "following inequality" is supposed to come from: it's certainly not obvious. At the end, you assume that $x_n > z_n$ (just before "Define:"), then use this to prove that $x_n > z_n$. You don't appear to conclude your conclusion at any point, or do anything that might constitute progress in that direction.
$endgroup$
– user3482749
Jan 10 at 16:43
$begingroup$
@user3482749 The inequality comes from $limleft(1+{xover n}right)^n = e^x$ and the fact that $left(1+{xover n}right)^n$ is monotonically increasing.
$endgroup$
– roman
Jan 10 at 16:46
$begingroup$
You are starting by assuming the inequality holds. Like, why would you assume what you are trying to prove? Or you meant to write that the inequality does not hold?
$endgroup$
– dezdichado
Jan 10 at 16:48
1
$begingroup$
@roman Say that then! The other problems all still hold. You've assumed the result at the start, and used it throughout.
$endgroup$
– user3482749
Jan 10 at 16:49
$begingroup$
It's hard to imagine a less silly mistake :(
$endgroup$
– roman
Jan 10 at 17:02
add a comment |
$begingroup$
Prove that for all $ninBbb N$:
$$
left(1+{1over n}right)^n > e^{1-{1over n}}
$$
Suppose the inequality holds.
Using the fact that:
$$
lim_{ntoinfty}left(1+{xover n}right)^n = e^x
$$
and:
$$
left(1+{xover n}right)^n le left(1+{xover n+1}right)^{n+1}
$$
put $x = {1-{1over n}}$ in the above:
$$
e^{1-{1over n}} > left(1+frac{1-{1over n}}{n}right)^n = \
= left(1+frac{n-1}{n^2}right)^n = y_n
$$
Let $x_n$:
$$
x_n = left(1+{1over n}right)^n
$$
Consider the fraction:
$$
frac{x_n}{y_n} = frac{left(1+{1over n}right)^n}{left(1+frac{n-1}{n^2}right)^n} \
=left(frac{n^2(n+1)}{n(n^2 + n - 1)}right)^n \
= left(frac{n^2 + n}{n^2 + n - 1}right)^n > 1
$$
Therefore $x_n > y_n$. But at the same time:
$$
e^{1-{1over n}} > y_n
$$
Thus:
$$
x_n > e^{1-{1over n}} > y_n
$$
Define:
$$
z_n = e^{1-{1over n}}
$$
Now if we take the limit of both sides then by squeeze theorem we get:
$$
lim_{ntoinfty}x_n gelim_{ntoinfty} z_n ge lim_{ntoinfty}y_n
$$
All three limits are equal to $e$ which shows that $z_n$ is squeezed between $x_n$ and $y_n$ meaning that $x_n > z_n$.
I'm not pretty sure the reasoning above is valid, so i'm kindly asking to verify it or point to the flaws. In case the above is wrong could you please show the way to prove the inequality? Thank you!
calculus sequences-and-series limits proof-verification inequality
asked Jan 10 at 16:39
romanroman
2,18521224
$endgroup$
Prove that for all $ninBbb N$:
$$
left(1+{1over n}right)^n > e^{1-{1over n}}
$$
Suppose the inequality holds.
Using the fact that:
$$
lim_{ntoinfty}left(1+{xover n}right)^n = e^x
$$
and:
$$
left(1+{xover n}right)^n le left(1+{xover n+1}right)^{n+1}
$$
put $x = {1-{1over n}}$ in the above:
$$
e^{1-{1over n}} > left(1+frac{1-{1over n}}{n}right)^n = \
= left(1+frac{n-1}{n^2}right)^n = y_n
$$
Let $x_n$:
$$
x_n = left(1+{1over n}right)^n
$$
Consider the fraction:
$$
frac{x_n}{y_n} = frac{left(1+{1over n}right)^n}{left(1+frac{n-1}{n^2}right)^n} \
=left(frac{n^2(n+1)}{n(n^2 + n - 1)}right)^n \
= left(frac{n^2 + n}{n^2 + n - 1}right)^n > 1
$$
Therefore $x_n > y_n$. But at the same time:
$$
e^{1-{1over n}} > y_n
$$
Thus:
$$
x_n > e^{1-{1over n}} > y_n
$$
Define:
$$
z_n = e^{1-{1over n}}
$$
Now if we take the limit of both sides then by squeeze theorem we get:
$$
lim_{ntoinfty}x_n gelim_{ntoinfty} z_n ge lim_{ntoinfty}y_n
$$
All three limits are equal to $e$ which shows that $z_n$ is squeezed between $x_n$ and $y_n$ meaning that $x_n > z_n$.
I'm not pretty sure the reasoning above is valid, so i'm kindly asking to verify it or point to the flaws. In case the above is wrong could you please show the way to prove the inequality? Thank you!
calculus sequences-and-series limits proof-verification inequality
calculus sequences-and-series limits proof-verification inequality
asked Jan 10 at 16:39
romanroman
2,18521224
asked Jan 10 at 16:39
romanroman
2,18521224
edited Jan 10 at 16:52
roman
asked Jan 10 at 16:39
romanroman
2,18521224
asked Jan 10 at 16:39
romanroman
2,18521224
asked Jan 10 at 16:39
romanroman
2,18521224
2,18521224
$begingroup$
Your proof appears to be backwards: the very first thing that you do is explicitly assume that your conclusion is true. I've actually no idea where the "following inequality" is supposed to come from: it's certainly not obvious. At the end, you assume that $x_n > z_n$ (just before "Define:"), then use this to prove that $x_n > z_n$. You don't appear to conclude your conclusion at any point, or do anything that might constitute progress in that direction.
$endgroup$
– user3482749
Jan 10 at 16:43
$begingroup$
@user3482749 The inequality comes from $limleft(1+{xover n}right)^n = e^x$ and the fact that $left(1+{xover n}right)^n$ is monotonically increasing.
$endgroup$
– roman
Jan 10 at 16:46
$begingroup$
You are starting by assuming the inequality holds. Like, why would you assume what you are trying to prove? Or you meant to write that the inequality does not hold?
$endgroup$
– dezdichado
Jan 10 at 16:48
1
$begingroup$
@roman Say that then! The other problems all still hold. You've assumed the result at the start, and used it throughout.
$endgroup$
– user3482749
Jan 10 at 16:49
$begingroup$
It's hard to imagine a less silly mistake :(
$endgroup$
– roman
Jan 10 at 17:02
add a comment |
$begingroup$
Your proof appears to be backwards: the very first thing that you do is explicitly assume that your conclusion is true. I've actually no idea where the "following inequality" is supposed to come from: it's certainly not obvious. At the end, you assume that $x_n > z_n$ (just before "Define:"), then use this to prove that $x_n > z_n$. You don't appear to conclude your conclusion at any point, or do anything that might constitute progress in that direction.
$endgroup$
– user3482749
Jan 10 at 16:43
$begingroup$
@user3482749 The inequality comes from $limleft(1+{xover n}right)^n = e^x$ and the fact that $left(1+{xover n}right)^n$ is monotonically increasing.
$endgroup$
– roman
Jan 10 at 16:46
$begingroup$
You are starting by assuming the inequality holds. Like, why would you assume what you are trying to prove? Or you meant to write that the inequality does not hold?
$endgroup$
– dezdichado
Jan 10 at 16:48
1
$begingroup$
@roman Say that then! The other problems all still hold. You've assumed the result at the start, and used it throughout.
$endgroup$
– user3482749
Jan 10 at 16:49
$begingroup$
It's hard to imagine a less silly mistake :(
$endgroup$
– roman
Jan 10 at 17:02
$begingroup$
Your proof appears to be backwards: the very first thing that you do is explicitly assume that your conclusion is true. I've actually no idea where the "following inequality" is supposed to come from: it's certainly not obvious. At the end, you assume that $x_n > z_n$ (just before "Define:"), then use this to prove that $x_n > z_n$. You don't appear to conclude your conclusion at any point, or do anything that might constitute progress in that direction.
$endgroup$
– user3482749
Jan 10 at 16:43
$begingroup$
Your proof appears to be backwards: the very first thing that you do is explicitly assume that your conclusion is true. I've actually no idea where the "following inequality" is supposed to come from: it's certainly not obvious. At the end, you assume that $x_n > z_n$ (just before "Define:"), then use this to prove that $x_n > z_n$. You don't appear to conclude your conclusion at any point, or do anything that might constitute progress in that direction.
$endgroup$
– user3482749
Jan 10 at 16:43
$begingroup$
@user3482749 The inequality comes from $limleft(1+{xover n}right)^n = e^x$ and the fact that $left(1+{xover n}right)^n$ is monotonically increasing.
$endgroup$
– roman
Jan 10 at 16:46
$begingroup$
@user3482749 The inequality comes from $limleft(1+{xover n}right)^n = e^x$ and the fact that $left(1+{xover n}right)^n$ is monotonically increasing.
$endgroup$
– roman
Jan 10 at 16:46
$begingroup$
You are starting by assuming the inequality holds. Like, why would you assume what you are trying to prove? Or you meant to write that the inequality does not hold?
$endgroup$
– dezdichado
Jan 10 at 16:48
$begingroup$
You are starting by assuming the inequality holds. Like, why would you assume what you are trying to prove? Or you meant to write that the inequality does not hold?
$endgroup$
– dezdichado
Jan 10 at 16:48
1
1
$begingroup$
@roman Say that then! The other problems all still hold. You've assumed the result at the start, and used it throughout.
$endgroup$
– user3482749
Jan 10 at 16:49
$begingroup$
@roman Say that then! The other problems all still hold. You've assumed the result at the start, and used it throughout.
$endgroup$
– user3482749
Jan 10 at 16:49
$begingroup$
It's hard to imagine a less silly mistake :(
$endgroup$
– roman
Jan 10 at 17:02
$begingroup$
It's hard to imagine a less silly mistake :(
$endgroup$
– roman
Jan 10 at 17:02
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Your flaw is that you are assuming the result to begin your proof. Here is a simple approach:
$$dfrac 1n +nlnleft(1+frac 1nright)>frac 1n+frac{n}{n+1} = dfrac{n^2+n+1}{n^2+n}>1.$$
Now, check that above is equivalent to your inequality. We used the well-known fact that:
$$ln(1+x)>dfrac{x}{x+1}$$
for $x>0.$
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
$endgroup$
add a comment |
$begingroup$
Why should the squeeze theorem imply that $x_n>z_n$? The theorem says that
if $x_n>z_n>y_n$ and $lim_{ntoinfty}x_n=l=lim_{ntoinfty}y_n$, then also $lim_{ntoinfty}z_n=l$.
However, it does not “reverse”. To see why, swap $z_n$ and $y_n$ in the statement above.
The inequality you have is equivalent to
$$
nlogBigl(1+frac{1}{n}Bigr)>1-frac{1}{n}
$$
or, setting $x=1/n$, to
$$
log(1+x)>x-x^2
$$
for $0<xle 1$. Consider $f(x)=log(1+x)-x+x^2$ defined over $(-1,1]$; then $f(0)=0$ and
$$
f'(x)=frac{2x^2+x}{1+x}>0
$$
proving that $f(x)>0$ for $0<xle1$.
answered Jan 10 at 17:13
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Here is a sketch of another approach: $frac{1}{1+1/n}le frac{1}{t}le 1$ as soon as $tin [1,1+1/n].$ Integrate across this inequality from $1$ to $1+1/n$ and find that $frac{1}{n+1}le ln(1+1/n)le 1/n.$ Now exponentiate and raise to the $n+1$ power, and simplify, and find that $left ( 1+1/n right )^{n}le ele left ( 1+1/n right )^{n+1}.$ The result follows easily from this.
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
$endgroup$
add a comment |
$begingroup$
Applying $ln$ to both sides, we see this holds iff
$$nln(1+1/n) > 1-1/n$$
for $nin mathbb N.$ This is the same as saying
$$frac{ln(1+1/n)-ln 1}{1/n}> 1-1/n.$$
By the MVT, the left side equals $1/c_n,$ where $c_nin (1,1+1/n).$ Since $1/c_n> 1/(1+1/n)$ and $1>(1+1/n)(1-1/n)$ for all $nin mathbb N,$ we're done.
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your flaw is that you are assuming the result to begin your proof. Here is a simple approach:
$$dfrac 1n +nlnleft(1+frac 1nright)>frac 1n+frac{n}{n+1} = dfrac{n^2+n+1}{n^2+n}>1.$$
Now, check that above is equivalent to your inequality. We used the well-known fact that:
$$ln(1+x)>dfrac{x}{x+1}$$
for $x>0.$
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
$endgroup$
add a comment |
$begingroup$
Your flaw is that you are assuming the result to begin your proof. Here is a simple approach:
$$dfrac 1n +nlnleft(1+frac 1nright)>frac 1n+frac{n}{n+1} = dfrac{n^2+n+1}{n^2+n}>1.$$
Now, check that above is equivalent to your inequality. We used the well-known fact that:
$$ln(1+x)>dfrac{x}{x+1}$$
for $x>0.$
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
$endgroup$
add a comment |
$begingroup$
Your flaw is that you are assuming the result to begin your proof. Here is a simple approach:
$$dfrac 1n +nlnleft(1+frac 1nright)>frac 1n+frac{n}{n+1} = dfrac{n^2+n+1}{n^2+n}>1.$$
Now, check that above is equivalent to your inequality. We used the well-known fact that:
$$ln(1+x)>dfrac{x}{x+1}$$
for $x>0.$
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
$endgroup$
Your flaw is that you are assuming the result to begin your proof. Here is a simple approach:
$$dfrac 1n +nlnleft(1+frac 1nright)>frac 1n+frac{n}{n+1} = dfrac{n^2+n+1}{n^2+n}>1.$$
Now, check that above is equivalent to your inequality. We used the well-known fact that:
$$ln(1+x)>dfrac{x}{x+1}$$
for $x>0.$
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
answered Jan 10 at 16:56
dezdichadodezdichado
6,3711929
6,3711929
add a comment |
add a comment |
$begingroup$
Why should the squeeze theorem imply that $x_n>z_n$? The theorem says that
if $x_n>z_n>y_n$ and $lim_{ntoinfty}x_n=l=lim_{ntoinfty}y_n$, then also $lim_{ntoinfty}z_n=l$.
However, it does not “reverse”. To see why, swap $z_n$ and $y_n$ in the statement above.
The inequality you have is equivalent to
$$
nlogBigl(1+frac{1}{n}Bigr)>1-frac{1}{n}
$$
or, setting $x=1/n$, to
$$
log(1+x)>x-x^2
$$
for $0<xle 1$. Consider $f(x)=log(1+x)-x+x^2$ defined over $(-1,1]$; then $f(0)=0$ and
$$
f'(x)=frac{2x^2+x}{1+x}>0
$$
proving that $f(x)>0$ for $0<xle1$.
answered Jan 10 at 17:13
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Why should the squeeze theorem imply that $x_n>z_n$? The theorem says that
if $x_n>z_n>y_n$ and $lim_{ntoinfty}x_n=l=lim_{ntoinfty}y_n$, then also $lim_{ntoinfty}z_n=l$.
However, it does not “reverse”. To see why, swap $z_n$ and $y_n$ in the statement above.
The inequality you have is equivalent to
$$
nlogBigl(1+frac{1}{n}Bigr)>1-frac{1}{n}
$$
or, setting $x=1/n$, to
$$
log(1+x)>x-x^2
$$
for $0<xle 1$. Consider $f(x)=log(1+x)-x+x^2$ defined over $(-1,1]$; then $f(0)=0$ and
$$
f'(x)=frac{2x^2+x}{1+x}>0
$$
proving that $f(x)>0$ for $0<xle1$.
answered Jan 10 at 17:13
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Why should the squeeze theorem imply that $x_n>z_n$? The theorem says that
if $x_n>z_n>y_n$ and $lim_{ntoinfty}x_n=l=lim_{ntoinfty}y_n$, then also $lim_{ntoinfty}z_n=l$.
However, it does not “reverse”. To see why, swap $z_n$ and $y_n$ in the statement above.
The inequality you have is equivalent to
$$
nlogBigl(1+frac{1}{n}Bigr)>1-frac{1}{n}
$$
or, setting $x=1/n$, to
$$
log(1+x)>x-x^2
$$
for $0<xle 1$. Consider $f(x)=log(1+x)-x+x^2$ defined over $(-1,1]$; then $f(0)=0$ and
$$
f'(x)=frac{2x^2+x}{1+x}>0
$$
proving that $f(x)>0$ for $0<xle1$.
answered Jan 10 at 17:13
egregegreg
181k1485203
$endgroup$
Why should the squeeze theorem imply that $x_n>z_n$? The theorem says that
if $x_n>z_n>y_n$ and $lim_{ntoinfty}x_n=l=lim_{ntoinfty}y_n$, then also $lim_{ntoinfty}z_n=l$.
However, it does not “reverse”. To see why, swap $z_n$ and $y_n$ in the statement above.
The inequality you have is equivalent to
$$
nlogBigl(1+frac{1}{n}Bigr)>1-frac{1}{n}
$$
or, setting $x=1/n$, to
$$
log(1+x)>x-x^2
$$
for $0<xle 1$. Consider $f(x)=log(1+x)-x+x^2$ defined over $(-1,1]$; then $f(0)=0$ and
$$
f'(x)=frac{2x^2+x}{1+x}>0
$$
proving that $f(x)>0$ for $0<xle1$.
answered Jan 10 at 17:13
egregegreg
181k1485203
answered Jan 10 at 17:13
egregegreg
181k1485203
answered Jan 10 at 17:13
egregegreg
181k1485203
answered Jan 10 at 17:13
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
Here is a sketch of another approach: $frac{1}{1+1/n}le frac{1}{t}le 1$ as soon as $tin [1,1+1/n].$ Integrate across this inequality from $1$ to $1+1/n$ and find that $frac{1}{n+1}le ln(1+1/n)le 1/n.$ Now exponentiate and raise to the $n+1$ power, and simplify, and find that $left ( 1+1/n right )^{n}le ele left ( 1+1/n right )^{n+1}.$ The result follows easily from this.
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
$endgroup$
add a comment |
$begingroup$
Here is a sketch of another approach: $frac{1}{1+1/n}le frac{1}{t}le 1$ as soon as $tin [1,1+1/n].$ Integrate across this inequality from $1$ to $1+1/n$ and find that $frac{1}{n+1}le ln(1+1/n)le 1/n.$ Now exponentiate and raise to the $n+1$ power, and simplify, and find that $left ( 1+1/n right )^{n}le ele left ( 1+1/n right )^{n+1}.$ The result follows easily from this.
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
$endgroup$
add a comment |
$begingroup$
Here is a sketch of another approach: $frac{1}{1+1/n}le frac{1}{t}le 1$ as soon as $tin [1,1+1/n].$ Integrate across this inequality from $1$ to $1+1/n$ and find that $frac{1}{n+1}le ln(1+1/n)le 1/n.$ Now exponentiate and raise to the $n+1$ power, and simplify, and find that $left ( 1+1/n right )^{n}le ele left ( 1+1/n right )^{n+1}.$ The result follows easily from this.
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
$endgroup$
Here is a sketch of another approach: $frac{1}{1+1/n}le frac{1}{t}le 1$ as soon as $tin [1,1+1/n].$ Integrate across this inequality from $1$ to $1+1/n$ and find that $frac{1}{n+1}le ln(1+1/n)le 1/n.$ Now exponentiate and raise to the $n+1$ power, and simplify, and find that $left ( 1+1/n right )^{n}le ele left ( 1+1/n right )^{n+1}.$ The result follows easily from this.
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
answered Jan 10 at 17:17
MatematletaMatematleta
10.7k2918
10.7k2918
add a comment |
add a comment |
$begingroup$
Applying $ln$ to both sides, we see this holds iff
$$nln(1+1/n) > 1-1/n$$
for $nin mathbb N.$ This is the same as saying
$$frac{ln(1+1/n)-ln 1}{1/n}> 1-1/n.$$
By the MVT, the left side equals $1/c_n,$ where $c_nin (1,1+1/n).$ Since $1/c_n> 1/(1+1/n)$ and $1>(1+1/n)(1-1/n)$ for all $nin mathbb N,$ we're done.
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
$endgroup$
add a comment |
$begingroup$
Applying $ln$ to both sides, we see this holds iff
$$nln(1+1/n) > 1-1/n$$
for $nin mathbb N.$ This is the same as saying
$$frac{ln(1+1/n)-ln 1}{1/n}> 1-1/n.$$
By the MVT, the left side equals $1/c_n,$ where $c_nin (1,1+1/n).$ Since $1/c_n> 1/(1+1/n)$ and $1>(1+1/n)(1-1/n)$ for all $nin mathbb N,$ we're done.
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
$endgroup$
add a comment |
$begingroup$
Applying $ln$ to both sides, we see this holds iff
$$nln(1+1/n) > 1-1/n$$
for $nin mathbb N.$ This is the same as saying
$$frac{ln(1+1/n)-ln 1}{1/n}> 1-1/n.$$
By the MVT, the left side equals $1/c_n,$ where $c_nin (1,1+1/n).$ Since $1/c_n> 1/(1+1/n)$ and $1>(1+1/n)(1-1/n)$ for all $nin mathbb N,$ we're done.
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
$endgroup$
Applying $ln$ to both sides, we see this holds iff
$$nln(1+1/n) > 1-1/n$$
for $nin mathbb N.$ This is the same as saying
$$frac{ln(1+1/n)-ln 1}{1/n}> 1-1/n.$$
By the MVT, the left side equals $1/c_n,$ where $c_nin (1,1+1/n).$ Since $1/c_n> 1/(1+1/n)$ and $1>(1+1/n)(1-1/n)$ for all $nin mathbb N,$ we're done.
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
answered Jan 11 at 0:13
zhw.zhw.
72.7k43175
72.7k43175
add a comment |
add a comment |
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$begingroup$
Your proof appears to be backwards: the very first thing that you do is explicitly assume that your conclusion is true. I've actually no idea where the "following inequality" is supposed to come from: it's certainly not obvious. At the end, you assume that $x_n > z_n$ (just before "Define:"), then use this to prove that $x_n > z_n$. You don't appear to conclude your conclusion at any point, or do anything that might constitute progress in that direction.
$endgroup$
– user3482749
Jan 10 at 16:43
$begingroup$
@user3482749 The inequality comes from $limleft(1+{xover n}right)^n = e^x$ and the fact that $left(1+{xover n}right)^n$ is monotonically increasing.
$endgroup$
– roman
Jan 10 at 16:46
$begingroup$
You are starting by assuming the inequality holds. Like, why would you assume what you are trying to prove? Or you meant to write that the inequality does not hold?
$endgroup$
– dezdichado
Jan 10 at 16:48
1
$begingroup$
@roman Say that then! The other problems all still hold. You've assumed the result at the start, and used it throughout.
$endgroup$
– user3482749
Jan 10 at 16:49
$begingroup$
It's hard to imagine a less silly mistake :(
$endgroup$
– roman
Jan 10 at 17:02