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Prove that lim $(sqrt{n^2+n}-n) = frac{1}{2}$
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Here's the question: Prove that lim $(sqrt{n^2+n}-n) = frac{1}{2}$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$left|sqrt{n^2+n}-n-frac{1}{2}right| < epsilon$
$Rightarrow left|frac{n}{sqrt{n^2+n}+n} - frac{1}{2}right| < epsilon$
$Rightarrow frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1} < epsilon$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}+1} > frac{1}{2} - epsilon = frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow sqrt{n} > frac{1-2 epsilon}{2}$
$Rightarrow n > frac{4 {epsilon}^2-4 epsilon +1}{4}$
real-analysis limits radicals
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
$endgroup$
add a comment |
$begingroup$
Here's the question: Prove that lim $(sqrt{n^2+n}-n) = frac{1}{2}$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$left|sqrt{n^2+n}-n-frac{1}{2}right| < epsilon$
$Rightarrow left|frac{n}{sqrt{n^2+n}+n} - frac{1}{2}right| < epsilon$
$Rightarrow frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1} < epsilon$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}+1} > frac{1}{2} - epsilon = frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow sqrt{n} > frac{1-2 epsilon}{2}$
$Rightarrow n > frac{4 {epsilon}^2-4 epsilon +1}{4}$
real-analysis limits radicals
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
$endgroup$
3
$begingroup$
You need the implications in the other direction, and $$frac{1}{sqrt{frac{1}{n}}} > frac{1-2epsilon}{2} notRightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2epsilon}{2}.$$
$endgroup$
– Daniel Fischer
May 6 '14 at 12:42
3
$begingroup$
Btw. unless you are explicitly asked to find $N$ for given $epsilon$ etc., you should more or less stop after reaching transforming $sqrt {n^2+n}-n$ to $frac1{1+sqrt{1+frac1n}}$. Then note $frac1nto 0$, hence $1+frac1nto 1$, hence $sqrt{1+frac1n}to 1$, hence $1+sqrt{1+frac1n}to 2$, hence $frac1{1+sqrt{1+frac1n}}tofrac12$.
$endgroup$
– Hagen von Eitzen
May 6 '14 at 12:48
add a comment |
$begingroup$
Here's the question: Prove that lim $(sqrt{n^2+n}-n) = frac{1}{2}$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$left|sqrt{n^2+n}-n-frac{1}{2}right| < epsilon$
$Rightarrow left|frac{n}{sqrt{n^2+n}+n} - frac{1}{2}right| < epsilon$
$Rightarrow frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1} < epsilon$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}+1} > frac{1}{2} - epsilon = frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow sqrt{n} > frac{1-2 epsilon}{2}$
$Rightarrow n > frac{4 {epsilon}^2-4 epsilon +1}{4}$
real-analysis limits radicals
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
$endgroup$
Here's the question: Prove that lim $(sqrt{n^2+n}-n) = frac{1}{2}$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$left|sqrt{n^2+n}-n-frac{1}{2}right| < epsilon$
$Rightarrow left|frac{n}{sqrt{n^2+n}+n} - frac{1}{2}right| < epsilon$
$Rightarrow frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1} < epsilon$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}+1} > frac{1}{2} - epsilon = frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow frac{1}{sqrt{frac{1}{n}}} > frac{1-2 epsilon}{2}$
$Rightarrow sqrt{n} > frac{1-2 epsilon}{2}$
$Rightarrow n > frac{4 {epsilon}^2-4 epsilon +1}{4}$
real-analysis limits radicals
real-analysis limits radicals
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
edited Jan 30 '16 at 4:14
Martin Sleziak
45k10122277
45k10122277
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
asked May 6 '14 at 12:39
Ryan StilesRyan Stiles
169126
169126
3
$begingroup$
You need the implications in the other direction, and $$frac{1}{sqrt{frac{1}{n}}} > frac{1-2epsilon}{2} notRightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2epsilon}{2}.$$
$endgroup$
– Daniel Fischer
May 6 '14 at 12:42
3
$begingroup$
Btw. unless you are explicitly asked to find $N$ for given $epsilon$ etc., you should more or less stop after reaching transforming $sqrt {n^2+n}-n$ to $frac1{1+sqrt{1+frac1n}}$. Then note $frac1nto 0$, hence $1+frac1nto 1$, hence $sqrt{1+frac1n}to 1$, hence $1+sqrt{1+frac1n}to 2$, hence $frac1{1+sqrt{1+frac1n}}tofrac12$.
$endgroup$
– Hagen von Eitzen
May 6 '14 at 12:48
add a comment |
3
$begingroup$
You need the implications in the other direction, and $$frac{1}{sqrt{frac{1}{n}}} > frac{1-2epsilon}{2} notRightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2epsilon}{2}.$$
$endgroup$
– Daniel Fischer
May 6 '14 at 12:42
3
$begingroup$
Btw. unless you are explicitly asked to find $N$ for given $epsilon$ etc., you should more or less stop after reaching transforming $sqrt {n^2+n}-n$ to $frac1{1+sqrt{1+frac1n}}$. Then note $frac1nto 0$, hence $1+frac1nto 1$, hence $sqrt{1+frac1n}to 1$, hence $1+sqrt{1+frac1n}to 2$, hence $frac1{1+sqrt{1+frac1n}}tofrac12$.
$endgroup$
– Hagen von Eitzen
May 6 '14 at 12:48
3
3
$begingroup$
You need the implications in the other direction, and $$frac{1}{sqrt{frac{1}{n}}} > frac{1-2epsilon}{2} notRightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2epsilon}{2}.$$
$endgroup$
– Daniel Fischer
May 6 '14 at 12:42
$begingroup$
You need the implications in the other direction, and $$frac{1}{sqrt{frac{1}{n}}} > frac{1-2epsilon}{2} notRightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2epsilon}{2}.$$
$endgroup$
– Daniel Fischer
May 6 '14 at 12:42
3
3
$begingroup$
Btw. unless you are explicitly asked to find $N$ for given $epsilon$ etc., you should more or less stop after reaching transforming $sqrt {n^2+n}-n$ to $frac1{1+sqrt{1+frac1n}}$. Then note $frac1nto 0$, hence $1+frac1nto 1$, hence $sqrt{1+frac1n}to 1$, hence $1+sqrt{1+frac1n}to 2$, hence $frac1{1+sqrt{1+frac1n}}tofrac12$.
$endgroup$
– Hagen von Eitzen
May 6 '14 at 12:48
$begingroup$
Btw. unless you are explicitly asked to find $N$ for given $epsilon$ etc., you should more or less stop after reaching transforming $sqrt {n^2+n}-n$ to $frac1{1+sqrt{1+frac1n}}$. Then note $frac1nto 0$, hence $1+frac1nto 1$, hence $sqrt{1+frac1n}to 1$, hence $1+sqrt{1+frac1n}to 2$, hence $frac1{1+sqrt{1+frac1n}}tofrac12$.
$endgroup$
– Hagen von Eitzen
May 6 '14 at 12:48
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
From $a^2-b^2=(a+b)(a-b)$ we have
$$sqrt{n^2+n}-n=frac{n^2+n-n^2}{sqrt{n^2+n}+n}=frac{n}{sqrt{n^2+n}+n}=
frac{1}{sqrt{1+frac{1}{n}}+1}$$
from which the result follows immediately.
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
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add a comment |
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Here is another solution. Remark that
$$
sqrt{n^2+n}-n=sqrt{n^2+n}-sqrt{n^2}=frac{(sqrt{n^2+n}-sqrt{n^2})(sqrt{n^2+n}+sqrt{n^2})}{sqrt{n^2+n}+sqrt{n^2}}
$$
Then
$$
sqrt{n^2+n}-n=frac{n}{sqrt{n^2+n}+n}=frac{1}{sqrt{1+frac{1}{n}}+1}.
$$
Since $limlimits_{nto infty} sqrt{1+frac{1}{n}}=1$, for $delta>0$, exists $N$ ($N>frac{1}{2delta+delta^2}$) such that $1<sqrt{1+frac{1}{n}}<1+delta$, for all $n>N$. Then
$$
2<sqrt{1+frac{1}{n}}+1<2+delta
$$
So
$$
frac{1}{2}>frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1}{2+delta}.
$$
Then
$$
frac{delta}{4}>frac{1}{2}-frac{1}{2+delta}>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0
$$
Let $varepsilon>0$, define $delta:=4varepsilon$. So, for $N>frac{1}{2delta+delta^2}=frac{1}{8varepsilon+16varepsilon^2}$,
$$
varepsilon>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0,
$$
i.e.,
$$
limlimits_{nto infty}sqrt{n^2+n}-n=limlimits_{nto infty}frac{1}{sqrt{1+frac{1}{n}}+1}=frac{1}{1+1}=frac{1}{2}
$$
answered May 6 '14 at 13:24
user73454user73454
501314
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add a comment |
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You can simply solve the inequality $frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1-2epsilon}{2}$and find pick N = $frac{(1-2epsilon)}{8epsilon^2}$
answered Feb 1 at 12:33
user556861user556861
311
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In general, be wary of the proofs by implications of that sort -- it is easy to miss a step, or to think an implication is an equivalence. Most of the time, what you want can be written as a succession of (in)equalities, but less tricky to handle.
For here, a way to handle this limit, if you have seen what equivalents or $o(cdot)$'s are: you can start by writing
$$
sqrt{n^2+n}-n = nsqrt{1+frac{1}{n}}-n = nleft(sqrt{1+frac{1}{n}} - 1right)
$$
and use the fact that
$$
(1+x)^alpha operatorname*{=}_{xto 0} 1+alpha x + o(x)
$$
to show that the term in the parentheses is equivalent to ("behaves like") $frac{1}{2n}$.
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
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Maybe the answer that gives the direct clue about what is going on is going through a proof that
$$lim_{n to +infty} (sqrt{n^2+n+c_2}-n)-(sqrt{n^2+n+c_1}-n)=0$$
or equally
$$lim_{n to +infty} sqrt{n^2+n+c_2}-sqrt{n^2+n+c_1}=0$$
Now write this to make it obvious as
$$lim_{n to +infty} frac{1}{sqrt{n^2+n+c_1}}(frac{sqrt{1+frac{1}{n}+frac{c_2}{n^2}}}{sqrt{1+frac{1}{n}+frac{c_1}{n^2}}}-1)=0$$
So we can pick whatever constant we want. We pick $frac{1}{4}$ and have
$$lim_{n to +infty} sqrt{n^2+n}-n=lim_{n to +infty} (sqrt{n^2+n+frac{1}{4}}-n)=$$ $$lim_{n to +infty} (sqrt{(n+frac{1}{2})^2}-n)=lim_{n to +infty} n+frac{1}{2}-n=frac{1}{2}$$
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Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)} - x
$$
where
$$
n,k in Bbb N \
a_k in Bbb R
$$
Consider the following expansion:
$$
a^n – b^n = (a – b)left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 +
cdots + ab^{n – 2} + b^{n – 1}right) tag1
$$
Define a function in the form:
$$
g(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)}
$$
Thus $f(x)$ may be rewritten as:
$$
f(x) = g(x) - x
$$
Now using $(1)$ we may rewrite $f(x)$ as:
$$
f(x) = frac{(g(x))^n-x^n}{sqrt[n]{(g(x))^{n-1}} +sqrt[n]{(g(x))^{n-2}}x + sqrt[n]{(g(x))^{n-3}}x^2 + cdots + sqrt[n]{g(x)}x^{n-2} +x^{n-1}} tag2
$$
Now taking the limit of $(2)$ one may obtain:
$$
begin{align*}
&lim_{xtoinfty}frac{x^{n-1}a_1 + x^{n-1}a_2+ cdots + x^{n-1}a_n }{x^{n-1}left(sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1right)} \
= &lim_{xtoinfty}frac{a_1 + a_2+ cdots + a_n}{sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1} \
= &frac{a_1 + a_2 + cdots + a_n}{n} tag3
end{align*}
$$
Or summarizing:
$$
boxed{lim_{xtoinfty}f(x) = frac{a_1 + a_2 + cdots + a_n}{n}}
$$
Lets now use that result in your limit:
$$
sqrt{n^2 + n} - n = sqrt{n(n+1)} - n = sqrt{(n+0)(n+1)} - n
$$
Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get:
$$
boxed{lim_{ntoinfty}x_n = frac{a_1 + a_2}{2} = frac{0 + 1}{2} = {1over 2}}
$$
As desired.
answered Feb 1 at 13:15
romanroman
2,50521226
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(Yet another solution - sorry!)
$$
left(1 + frac1{2n}right)^2 = 1 + frac1n + frac1{4n^2} > 1 + frac1n,
$$
therefore
$$
sqrt{1 + frac1n} < 1 + frac1{2n},
$$
therefore
$$
sqrt{n^2 + n} - n = nleft(sqrt{1 + frac1n} - 1right) < frac12,
$$
but also
$$
sqrt{n^2 + n} - n = frac1{sqrt{1 + frac1n} + 1} > frac1{2left(1 + frac1{4n}right)}
> frac{1 - frac1{4n}}{2} = frac12 - frac1{8n},
$$
therefore
$$
frac12 - epsilon < sqrt{n^2 + n} - n < frac12
quad left(epsilon > 0, n geqslant frac1{8epsilon}right).
$$
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
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7 Answers
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7 Answers
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$begingroup$
From $a^2-b^2=(a+b)(a-b)$ we have
$$sqrt{n^2+n}-n=frac{n^2+n-n^2}{sqrt{n^2+n}+n}=frac{n}{sqrt{n^2+n}+n}=
frac{1}{sqrt{1+frac{1}{n}}+1}$$
from which the result follows immediately.
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
$endgroup$
add a comment |
$begingroup$
From $a^2-b^2=(a+b)(a-b)$ we have
$$sqrt{n^2+n}-n=frac{n^2+n-n^2}{sqrt{n^2+n}+n}=frac{n}{sqrt{n^2+n}+n}=
frac{1}{sqrt{1+frac{1}{n}}+1}$$
from which the result follows immediately.
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
$endgroup$
add a comment |
$begingroup$
From $a^2-b^2=(a+b)(a-b)$ we have
$$sqrt{n^2+n}-n=frac{n^2+n-n^2}{sqrt{n^2+n}+n}=frac{n}{sqrt{n^2+n}+n}=
frac{1}{sqrt{1+frac{1}{n}}+1}$$
from which the result follows immediately.
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
$endgroup$
From $a^2-b^2=(a+b)(a-b)$ we have
$$sqrt{n^2+n}-n=frac{n^2+n-n^2}{sqrt{n^2+n}+n}=frac{n}{sqrt{n^2+n}+n}=
frac{1}{sqrt{1+frac{1}{n}}+1}$$
from which the result follows immediately.
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
answered May 6 '14 at 13:19
Matt L.Matt L.
9,034922
9,034922
add a comment |
add a comment |
$begingroup$
Here is another solution. Remark that
$$
sqrt{n^2+n}-n=sqrt{n^2+n}-sqrt{n^2}=frac{(sqrt{n^2+n}-sqrt{n^2})(sqrt{n^2+n}+sqrt{n^2})}{sqrt{n^2+n}+sqrt{n^2}}
$$
Then
$$
sqrt{n^2+n}-n=frac{n}{sqrt{n^2+n}+n}=frac{1}{sqrt{1+frac{1}{n}}+1}.
$$
Since $limlimits_{nto infty} sqrt{1+frac{1}{n}}=1$, for $delta>0$, exists $N$ ($N>frac{1}{2delta+delta^2}$) such that $1<sqrt{1+frac{1}{n}}<1+delta$, for all $n>N$. Then
$$
2<sqrt{1+frac{1}{n}}+1<2+delta
$$
So
$$
frac{1}{2}>frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1}{2+delta}.
$$
Then
$$
frac{delta}{4}>frac{1}{2}-frac{1}{2+delta}>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0
$$
Let $varepsilon>0$, define $delta:=4varepsilon$. So, for $N>frac{1}{2delta+delta^2}=frac{1}{8varepsilon+16varepsilon^2}$,
$$
varepsilon>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0,
$$
i.e.,
$$
limlimits_{nto infty}sqrt{n^2+n}-n=limlimits_{nto infty}frac{1}{sqrt{1+frac{1}{n}}+1}=frac{1}{1+1}=frac{1}{2}
$$
answered May 6 '14 at 13:24
user73454user73454
501314
$endgroup$
add a comment |
$begingroup$
Here is another solution. Remark that
$$
sqrt{n^2+n}-n=sqrt{n^2+n}-sqrt{n^2}=frac{(sqrt{n^2+n}-sqrt{n^2})(sqrt{n^2+n}+sqrt{n^2})}{sqrt{n^2+n}+sqrt{n^2}}
$$
Then
$$
sqrt{n^2+n}-n=frac{n}{sqrt{n^2+n}+n}=frac{1}{sqrt{1+frac{1}{n}}+1}.
$$
Since $limlimits_{nto infty} sqrt{1+frac{1}{n}}=1$, for $delta>0$, exists $N$ ($N>frac{1}{2delta+delta^2}$) such that $1<sqrt{1+frac{1}{n}}<1+delta$, for all $n>N$. Then
$$
2<sqrt{1+frac{1}{n}}+1<2+delta
$$
So
$$
frac{1}{2}>frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1}{2+delta}.
$$
Then
$$
frac{delta}{4}>frac{1}{2}-frac{1}{2+delta}>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0
$$
Let $varepsilon>0$, define $delta:=4varepsilon$. So, for $N>frac{1}{2delta+delta^2}=frac{1}{8varepsilon+16varepsilon^2}$,
$$
varepsilon>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0,
$$
i.e.,
$$
limlimits_{nto infty}sqrt{n^2+n}-n=limlimits_{nto infty}frac{1}{sqrt{1+frac{1}{n}}+1}=frac{1}{1+1}=frac{1}{2}
$$
answered May 6 '14 at 13:24
user73454user73454
501314
$endgroup$
add a comment |
$begingroup$
Here is another solution. Remark that
$$
sqrt{n^2+n}-n=sqrt{n^2+n}-sqrt{n^2}=frac{(sqrt{n^2+n}-sqrt{n^2})(sqrt{n^2+n}+sqrt{n^2})}{sqrt{n^2+n}+sqrt{n^2}}
$$
Then
$$
sqrt{n^2+n}-n=frac{n}{sqrt{n^2+n}+n}=frac{1}{sqrt{1+frac{1}{n}}+1}.
$$
Since $limlimits_{nto infty} sqrt{1+frac{1}{n}}=1$, for $delta>0$, exists $N$ ($N>frac{1}{2delta+delta^2}$) such that $1<sqrt{1+frac{1}{n}}<1+delta$, for all $n>N$. Then
$$
2<sqrt{1+frac{1}{n}}+1<2+delta
$$
So
$$
frac{1}{2}>frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1}{2+delta}.
$$
Then
$$
frac{delta}{4}>frac{1}{2}-frac{1}{2+delta}>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0
$$
Let $varepsilon>0$, define $delta:=4varepsilon$. So, for $N>frac{1}{2delta+delta^2}=frac{1}{8varepsilon+16varepsilon^2}$,
$$
varepsilon>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0,
$$
i.e.,
$$
limlimits_{nto infty}sqrt{n^2+n}-n=limlimits_{nto infty}frac{1}{sqrt{1+frac{1}{n}}+1}=frac{1}{1+1}=frac{1}{2}
$$
answered May 6 '14 at 13:24
user73454user73454
501314
$endgroup$
Here is another solution. Remark that
$$
sqrt{n^2+n}-n=sqrt{n^2+n}-sqrt{n^2}=frac{(sqrt{n^2+n}-sqrt{n^2})(sqrt{n^2+n}+sqrt{n^2})}{sqrt{n^2+n}+sqrt{n^2}}
$$
Then
$$
sqrt{n^2+n}-n=frac{n}{sqrt{n^2+n}+n}=frac{1}{sqrt{1+frac{1}{n}}+1}.
$$
Since $limlimits_{nto infty} sqrt{1+frac{1}{n}}=1$, for $delta>0$, exists $N$ ($N>frac{1}{2delta+delta^2}$) such that $1<sqrt{1+frac{1}{n}}<1+delta$, for all $n>N$. Then
$$
2<sqrt{1+frac{1}{n}}+1<2+delta
$$
So
$$
frac{1}{2}>frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1}{2+delta}.
$$
Then
$$
frac{delta}{4}>frac{1}{2}-frac{1}{2+delta}>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0
$$
Let $varepsilon>0$, define $delta:=4varepsilon$. So, for $N>frac{1}{2delta+delta^2}=frac{1}{8varepsilon+16varepsilon^2}$,
$$
varepsilon>frac{1}{2}-frac{1}{sqrt{1+frac{1}{n}}+1}>0,
$$
i.e.,
$$
limlimits_{nto infty}sqrt{n^2+n}-n=limlimits_{nto infty}frac{1}{sqrt{1+frac{1}{n}}+1}=frac{1}{1+1}=frac{1}{2}
$$
answered May 6 '14 at 13:24
user73454user73454
501314
answered May 6 '14 at 13:24
user73454user73454
501314
answered May 6 '14 at 13:24
user73454user73454
501314
answered May 6 '14 at 13:24
user73454user73454
501314
501314
add a comment |
add a comment |
$begingroup$
You can simply solve the inequality $frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1-2epsilon}{2}$and find pick N = $frac{(1-2epsilon)}{8epsilon^2}$
answered Feb 1 at 12:33
user556861user556861
311
$endgroup$
add a comment |
$begingroup$
You can simply solve the inequality $frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1-2epsilon}{2}$and find pick N = $frac{(1-2epsilon)}{8epsilon^2}$
answered Feb 1 at 12:33
user556861user556861
311
$endgroup$
add a comment |
$begingroup$
You can simply solve the inequality $frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1-2epsilon}{2}$and find pick N = $frac{(1-2epsilon)}{8epsilon^2}$
answered Feb 1 at 12:33
user556861user556861
311
$endgroup$
You can simply solve the inequality $frac{1}{sqrt{1+frac{1}{n}}+1}>frac{1-2epsilon}{2}$and find pick N = $frac{(1-2epsilon)}{8epsilon^2}$
answered Feb 1 at 12:33
user556861user556861
311
answered Feb 1 at 12:33
user556861user556861
311
answered Feb 1 at 12:33
user556861user556861
311
answered Feb 1 at 12:33
user556861user556861
311
311
add a comment |
add a comment |
$begingroup$
In general, be wary of the proofs by implications of that sort -- it is easy to miss a step, or to think an implication is an equivalence. Most of the time, what you want can be written as a succession of (in)equalities, but less tricky to handle.
For here, a way to handle this limit, if you have seen what equivalents or $o(cdot)$'s are: you can start by writing
$$
sqrt{n^2+n}-n = nsqrt{1+frac{1}{n}}-n = nleft(sqrt{1+frac{1}{n}} - 1right)
$$
and use the fact that
$$
(1+x)^alpha operatorname*{=}_{xto 0} 1+alpha x + o(x)
$$
to show that the term in the parentheses is equivalent to ("behaves like") $frac{1}{2n}$.
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
$endgroup$
add a comment |
$begingroup$
In general, be wary of the proofs by implications of that sort -- it is easy to miss a step, or to think an implication is an equivalence. Most of the time, what you want can be written as a succession of (in)equalities, but less tricky to handle.
For here, a way to handle this limit, if you have seen what equivalents or $o(cdot)$'s are: you can start by writing
$$
sqrt{n^2+n}-n = nsqrt{1+frac{1}{n}}-n = nleft(sqrt{1+frac{1}{n}} - 1right)
$$
and use the fact that
$$
(1+x)^alpha operatorname*{=}_{xto 0} 1+alpha x + o(x)
$$
to show that the term in the parentheses is equivalent to ("behaves like") $frac{1}{2n}$.
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
$endgroup$
add a comment |
$begingroup$
In general, be wary of the proofs by implications of that sort -- it is easy to miss a step, or to think an implication is an equivalence. Most of the time, what you want can be written as a succession of (in)equalities, but less tricky to handle.
For here, a way to handle this limit, if you have seen what equivalents or $o(cdot)$'s are: you can start by writing
$$
sqrt{n^2+n}-n = nsqrt{1+frac{1}{n}}-n = nleft(sqrt{1+frac{1}{n}} - 1right)
$$
and use the fact that
$$
(1+x)^alpha operatorname*{=}_{xto 0} 1+alpha x + o(x)
$$
to show that the term in the parentheses is equivalent to ("behaves like") $frac{1}{2n}$.
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
$endgroup$
In general, be wary of the proofs by implications of that sort -- it is easy to miss a step, or to think an implication is an equivalence. Most of the time, what you want can be written as a succession of (in)equalities, but less tricky to handle.
For here, a way to handle this limit, if you have seen what equivalents or $o(cdot)$'s are: you can start by writing
$$
sqrt{n^2+n}-n = nsqrt{1+frac{1}{n}}-n = nleft(sqrt{1+frac{1}{n}} - 1right)
$$
and use the fact that
$$
(1+x)^alpha operatorname*{=}_{xto 0} 1+alpha x + o(x)
$$
to show that the term in the parentheses is equivalent to ("behaves like") $frac{1}{2n}$.
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
answered May 6 '14 at 12:50
Clement C.Clement C.
51.1k34093
51.1k34093
add a comment |
add a comment |
$begingroup$
Maybe the answer that gives the direct clue about what is going on is going through a proof that
$$lim_{n to +infty} (sqrt{n^2+n+c_2}-n)-(sqrt{n^2+n+c_1}-n)=0$$
or equally
$$lim_{n to +infty} sqrt{n^2+n+c_2}-sqrt{n^2+n+c_1}=0$$
Now write this to make it obvious as
$$lim_{n to +infty} frac{1}{sqrt{n^2+n+c_1}}(frac{sqrt{1+frac{1}{n}+frac{c_2}{n^2}}}{sqrt{1+frac{1}{n}+frac{c_1}{n^2}}}-1)=0$$
So we can pick whatever constant we want. We pick $frac{1}{4}$ and have
$$lim_{n to +infty} sqrt{n^2+n}-n=lim_{n to +infty} (sqrt{n^2+n+frac{1}{4}}-n)=$$ $$lim_{n to +infty} (sqrt{(n+frac{1}{2})^2}-n)=lim_{n to +infty} n+frac{1}{2}-n=frac{1}{2}$$
$endgroup$
add a comment |
$begingroup$
Maybe the answer that gives the direct clue about what is going on is going through a proof that
$$lim_{n to +infty} (sqrt{n^2+n+c_2}-n)-(sqrt{n^2+n+c_1}-n)=0$$
or equally
$$lim_{n to +infty} sqrt{n^2+n+c_2}-sqrt{n^2+n+c_1}=0$$
Now write this to make it obvious as
$$lim_{n to +infty} frac{1}{sqrt{n^2+n+c_1}}(frac{sqrt{1+frac{1}{n}+frac{c_2}{n^2}}}{sqrt{1+frac{1}{n}+frac{c_1}{n^2}}}-1)=0$$
So we can pick whatever constant we want. We pick $frac{1}{4}$ and have
$$lim_{n to +infty} sqrt{n^2+n}-n=lim_{n to +infty} (sqrt{n^2+n+frac{1}{4}}-n)=$$ $$lim_{n to +infty} (sqrt{(n+frac{1}{2})^2}-n)=lim_{n to +infty} n+frac{1}{2}-n=frac{1}{2}$$
$endgroup$
add a comment |
$begingroup$
Maybe the answer that gives the direct clue about what is going on is going through a proof that
$$lim_{n to +infty} (sqrt{n^2+n+c_2}-n)-(sqrt{n^2+n+c_1}-n)=0$$
or equally
$$lim_{n to +infty} sqrt{n^2+n+c_2}-sqrt{n^2+n+c_1}=0$$
Now write this to make it obvious as
$$lim_{n to +infty} frac{1}{sqrt{n^2+n+c_1}}(frac{sqrt{1+frac{1}{n}+frac{c_2}{n^2}}}{sqrt{1+frac{1}{n}+frac{c_1}{n^2}}}-1)=0$$
So we can pick whatever constant we want. We pick $frac{1}{4}$ and have
$$lim_{n to +infty} sqrt{n^2+n}-n=lim_{n to +infty} (sqrt{n^2+n+frac{1}{4}}-n)=$$ $$lim_{n to +infty} (sqrt{(n+frac{1}{2})^2}-n)=lim_{n to +infty} n+frac{1}{2}-n=frac{1}{2}$$
$endgroup$
Maybe the answer that gives the direct clue about what is going on is going through a proof that
$$lim_{n to +infty} (sqrt{n^2+n+c_2}-n)-(sqrt{n^2+n+c_1}-n)=0$$
or equally
$$lim_{n to +infty} sqrt{n^2+n+c_2}-sqrt{n^2+n+c_1}=0$$
Now write this to make it obvious as
$$lim_{n to +infty} frac{1}{sqrt{n^2+n+c_1}}(frac{sqrt{1+frac{1}{n}+frac{c_2}{n^2}}}{sqrt{1+frac{1}{n}+frac{c_1}{n^2}}}-1)=0$$
So we can pick whatever constant we want. We pick $frac{1}{4}$ and have
$$lim_{n to +infty} sqrt{n^2+n}-n=lim_{n to +infty} (sqrt{n^2+n+frac{1}{4}}-n)=$$ $$lim_{n to +infty} (sqrt{(n+frac{1}{2})^2}-n)=lim_{n to +infty} n+frac{1}{2}-n=frac{1}{2}$$
answered Apr 7 '18 at 19:29
user318107
add a comment |
add a comment |
$begingroup$
Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)} - x
$$
where
$$
n,k in Bbb N \
a_k in Bbb R
$$
Consider the following expansion:
$$
a^n – b^n = (a – b)left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 +
cdots + ab^{n – 2} + b^{n – 1}right) tag1
$$
Define a function in the form:
$$
g(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)}
$$
Thus $f(x)$ may be rewritten as:
$$
f(x) = g(x) - x
$$
Now using $(1)$ we may rewrite $f(x)$ as:
$$
f(x) = frac{(g(x))^n-x^n}{sqrt[n]{(g(x))^{n-1}} +sqrt[n]{(g(x))^{n-2}}x + sqrt[n]{(g(x))^{n-3}}x^2 + cdots + sqrt[n]{g(x)}x^{n-2} +x^{n-1}} tag2
$$
Now taking the limit of $(2)$ one may obtain:
$$
begin{align*}
&lim_{xtoinfty}frac{x^{n-1}a_1 + x^{n-1}a_2+ cdots + x^{n-1}a_n }{x^{n-1}left(sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1right)} \
= &lim_{xtoinfty}frac{a_1 + a_2+ cdots + a_n}{sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1} \
= &frac{a_1 + a_2 + cdots + a_n}{n} tag3
end{align*}
$$
Or summarizing:
$$
boxed{lim_{xtoinfty}f(x) = frac{a_1 + a_2 + cdots + a_n}{n}}
$$
Lets now use that result in your limit:
$$
sqrt{n^2 + n} - n = sqrt{n(n+1)} - n = sqrt{(n+0)(n+1)} - n
$$
Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get:
$$
boxed{lim_{ntoinfty}x_n = frac{a_1 + a_2}{2} = frac{0 + 1}{2} = {1over 2}}
$$
As desired.
answered Feb 1 at 13:15
romanroman
2,50521226
$endgroup$
add a comment |
$begingroup$
Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)} - x
$$
where
$$
n,k in Bbb N \
a_k in Bbb R
$$
Consider the following expansion:
$$
a^n – b^n = (a – b)left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 +
cdots + ab^{n – 2} + b^{n – 1}right) tag1
$$
Define a function in the form:
$$
g(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)}
$$
Thus $f(x)$ may be rewritten as:
$$
f(x) = g(x) - x
$$
Now using $(1)$ we may rewrite $f(x)$ as:
$$
f(x) = frac{(g(x))^n-x^n}{sqrt[n]{(g(x))^{n-1}} +sqrt[n]{(g(x))^{n-2}}x + sqrt[n]{(g(x))^{n-3}}x^2 + cdots + sqrt[n]{g(x)}x^{n-2} +x^{n-1}} tag2
$$
Now taking the limit of $(2)$ one may obtain:
$$
begin{align*}
&lim_{xtoinfty}frac{x^{n-1}a_1 + x^{n-1}a_2+ cdots + x^{n-1}a_n }{x^{n-1}left(sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1right)} \
= &lim_{xtoinfty}frac{a_1 + a_2+ cdots + a_n}{sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1} \
= &frac{a_1 + a_2 + cdots + a_n}{n} tag3
end{align*}
$$
Or summarizing:
$$
boxed{lim_{xtoinfty}f(x) = frac{a_1 + a_2 + cdots + a_n}{n}}
$$
Lets now use that result in your limit:
$$
sqrt{n^2 + n} - n = sqrt{n(n+1)} - n = sqrt{(n+0)(n+1)} - n
$$
Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get:
$$
boxed{lim_{ntoinfty}x_n = frac{a_1 + a_2}{2} = frac{0 + 1}{2} = {1over 2}}
$$
As desired.
answered Feb 1 at 13:15
romanroman
2,50521226
$endgroup$
add a comment |
$begingroup$
Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)} - x
$$
where
$$
n,k in Bbb N \
a_k in Bbb R
$$
Consider the following expansion:
$$
a^n – b^n = (a – b)left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 +
cdots + ab^{n – 2} + b^{n – 1}right) tag1
$$
Define a function in the form:
$$
g(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)}
$$
Thus $f(x)$ may be rewritten as:
$$
f(x) = g(x) - x
$$
Now using $(1)$ we may rewrite $f(x)$ as:
$$
f(x) = frac{(g(x))^n-x^n}{sqrt[n]{(g(x))^{n-1}} +sqrt[n]{(g(x))^{n-2}}x + sqrt[n]{(g(x))^{n-3}}x^2 + cdots + sqrt[n]{g(x)}x^{n-2} +x^{n-1}} tag2
$$
Now taking the limit of $(2)$ one may obtain:
$$
begin{align*}
&lim_{xtoinfty}frac{x^{n-1}a_1 + x^{n-1}a_2+ cdots + x^{n-1}a_n }{x^{n-1}left(sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1right)} \
= &lim_{xtoinfty}frac{a_1 + a_2+ cdots + a_n}{sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1} \
= &frac{a_1 + a_2 + cdots + a_n}{n} tag3
end{align*}
$$
Or summarizing:
$$
boxed{lim_{xtoinfty}f(x) = frac{a_1 + a_2 + cdots + a_n}{n}}
$$
Lets now use that result in your limit:
$$
sqrt{n^2 + n} - n = sqrt{n(n+1)} - n = sqrt{(n+0)(n+1)} - n
$$
Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get:
$$
boxed{lim_{ntoinfty}x_n = frac{a_1 + a_2}{2} = frac{0 + 1}{2} = {1over 2}}
$$
As desired.
answered Feb 1 at 13:15
romanroman
2,50521226
$endgroup$
Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form:
$$
f(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)} - x
$$
where
$$
n,k in Bbb N \
a_k in Bbb R
$$
Consider the following expansion:
$$
a^n – b^n = (a – b)left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 +
cdots + ab^{n – 2} + b^{n – 1}right) tag1
$$
Define a function in the form:
$$
g(x) = sqrt[n]{(x + a_1)(x+a_2)cdots (x+a_n)}
$$
Thus $f(x)$ may be rewritten as:
$$
f(x) = g(x) - x
$$
Now using $(1)$ we may rewrite $f(x)$ as:
$$
f(x) = frac{(g(x))^n-x^n}{sqrt[n]{(g(x))^{n-1}} +sqrt[n]{(g(x))^{n-2}}x + sqrt[n]{(g(x))^{n-3}}x^2 + cdots + sqrt[n]{g(x)}x^{n-2} +x^{n-1}} tag2
$$
Now taking the limit of $(2)$ one may obtain:
$$
begin{align*}
&lim_{xtoinfty}frac{x^{n-1}a_1 + x^{n-1}a_2+ cdots + x^{n-1}a_n }{x^{n-1}left(sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1right)} \
= &lim_{xtoinfty}frac{a_1 + a_2+ cdots + a_n}{sqrt[n]{1+oleft({1over x}right)+cdots} + sqrt[n]{1+oleft({1over x}right) +cdots} +cdots + 1} \
= &frac{a_1 + a_2 + cdots + a_n}{n} tag3
end{align*}
$$
Or summarizing:
$$
boxed{lim_{xtoinfty}f(x) = frac{a_1 + a_2 + cdots + a_n}{n}}
$$
Lets now use that result in your limit:
$$
sqrt{n^2 + n} - n = sqrt{n(n+1)} - n = sqrt{(n+0)(n+1)} - n
$$
Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get:
$$
boxed{lim_{ntoinfty}x_n = frac{a_1 + a_2}{2} = frac{0 + 1}{2} = {1over 2}}
$$
As desired.
answered Feb 1 at 13:15
romanroman
2,50521226
edited Feb 1 at 13:22
answered Feb 1 at 13:15
romanroman
2,50521226
answered Feb 1 at 13:15
romanroman
2,50521226
answered Feb 1 at 13:15
romanroman
2,50521226
2,50521226
add a comment |
add a comment |
$begingroup$
(Yet another solution - sorry!)
$$
left(1 + frac1{2n}right)^2 = 1 + frac1n + frac1{4n^2} > 1 + frac1n,
$$
therefore
$$
sqrt{1 + frac1n} < 1 + frac1{2n},
$$
therefore
$$
sqrt{n^2 + n} - n = nleft(sqrt{1 + frac1n} - 1right) < frac12,
$$
but also
$$
sqrt{n^2 + n} - n = frac1{sqrt{1 + frac1n} + 1} > frac1{2left(1 + frac1{4n}right)}
> frac{1 - frac1{4n}}{2} = frac12 - frac1{8n},
$$
therefore
$$
frac12 - epsilon < sqrt{n^2 + n} - n < frac12
quad left(epsilon > 0, n geqslant frac1{8epsilon}right).
$$
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
$endgroup$
add a comment |
$begingroup$
(Yet another solution - sorry!)
$$
left(1 + frac1{2n}right)^2 = 1 + frac1n + frac1{4n^2} > 1 + frac1n,
$$
therefore
$$
sqrt{1 + frac1n} < 1 + frac1{2n},
$$
therefore
$$
sqrt{n^2 + n} - n = nleft(sqrt{1 + frac1n} - 1right) < frac12,
$$
but also
$$
sqrt{n^2 + n} - n = frac1{sqrt{1 + frac1n} + 1} > frac1{2left(1 + frac1{4n}right)}
> frac{1 - frac1{4n}}{2} = frac12 - frac1{8n},
$$
therefore
$$
frac12 - epsilon < sqrt{n^2 + n} - n < frac12
quad left(epsilon > 0, n geqslant frac1{8epsilon}right).
$$
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
$endgroup$
add a comment |
$begingroup$
(Yet another solution - sorry!)
$$
left(1 + frac1{2n}right)^2 = 1 + frac1n + frac1{4n^2} > 1 + frac1n,
$$
therefore
$$
sqrt{1 + frac1n} < 1 + frac1{2n},
$$
therefore
$$
sqrt{n^2 + n} - n = nleft(sqrt{1 + frac1n} - 1right) < frac12,
$$
but also
$$
sqrt{n^2 + n} - n = frac1{sqrt{1 + frac1n} + 1} > frac1{2left(1 + frac1{4n}right)}
> frac{1 - frac1{4n}}{2} = frac12 - frac1{8n},
$$
therefore
$$
frac12 - epsilon < sqrt{n^2 + n} - n < frac12
quad left(epsilon > 0, n geqslant frac1{8epsilon}right).
$$
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
$endgroup$
(Yet another solution - sorry!)
$$
left(1 + frac1{2n}right)^2 = 1 + frac1n + frac1{4n^2} > 1 + frac1n,
$$
therefore
$$
sqrt{1 + frac1n} < 1 + frac1{2n},
$$
therefore
$$
sqrt{n^2 + n} - n = nleft(sqrt{1 + frac1n} - 1right) < frac12,
$$
but also
$$
sqrt{n^2 + n} - n = frac1{sqrt{1 + frac1n} + 1} > frac1{2left(1 + frac1{4n}right)}
> frac{1 - frac1{4n}}{2} = frac12 - frac1{8n},
$$
therefore
$$
frac12 - epsilon < sqrt{n^2 + n} - n < frac12
quad left(epsilon > 0, n geqslant frac1{8epsilon}right).
$$
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
answered Feb 1 at 14:06
Calum GilhooleyCalum Gilhooley
5,119730
5,119730
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3
$begingroup$
You need the implications in the other direction, and $$frac{1}{sqrt{frac{1}{n}}} > frac{1-2epsilon}{2} notRightarrow frac{1}{sqrt{1+frac{1}{n}}} > frac{1-2epsilon}{2}.$$
$endgroup$
– Daniel Fischer
May 6 '14 at 12:42
3
$begingroup$
Btw. unless you are explicitly asked to find $N$ for given $epsilon$ etc., you should more or less stop after reaching transforming $sqrt {n^2+n}-n$ to $frac1{1+sqrt{1+frac1n}}$. Then note $frac1nto 0$, hence $1+frac1nto 1$, hence $sqrt{1+frac1n}to 1$, hence $1+sqrt{1+frac1n}to 2$, hence $frac1{1+sqrt{1+frac1n}}tofrac12$.
$endgroup$
– Hagen von Eitzen
May 6 '14 at 12:48