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How do I compute this limit? $$ \lim_{n \to \infty} \frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n - \left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n }{ 2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n - \left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n - \left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n } $$

I think I got the correct limit by using fast converging limits to $e$. In particular I used truncated Taylor series for the sqrt and 4th root. Or squares and bisquares.

Example

$(1+1/2n)^{2n}$ Becomes $(1 + 1/n + 1/4n^2)^n.$

In combination with l'hopital it gives me the answer.

But I guess that is not a very good (fast) method.

mick
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    what are your thoughts? – tired Nov 04 '15 at 00:45
  • Close votes ? Really ? – mick Nov 04 '15 at 01:06
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    probably you have in the denominator a $ 2( 1+ \frac{1}{n}+ \frac{1}{n^2} )^n $, otherwise the answers solve you this problem – clark Nov 04 '15 at 01:13
  • Thanks Clark !! A typo sorry. – mick Nov 04 '15 at 01:21
  • I fail to see how this is " off topic " :/ Imho too many close votes these days ... Or is it just me ? – mick Nov 04 '15 at 05:46
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    Taylor formula at order $1$ on the function $f_n(x)=(1+1/n+x)^n$ yields readily, using the shorthands $a=f_n(0)\in\Theta(1)$ and $b=f'_n(0)\in\Theta(n)$, that the ratio is $$\frac{a+b/n^2-(a-b/n^2)+o(1/n)}{2(a+b/n^2)-(a-b/n^2)-a+o(1/n)}=\frac{2b/n^2+o(1/n)}{3b/n^2+o(1/n^2)}\to\frac23.$$ Note that exact asymptotics of $a$ and $b$, not to mention the exact limit of $a$, are irrelevant. – Did Nov 04 '15 at 07:10
  • @Did In case you saw it, I just wrote you a comment about your post: I deleted my comment, because you were right, and I was wrong... sigh – peter a g Nov 04 '15 at 14:33

6 Answers6

2

Hint: First show $(1+1/n +o(1/n))^n \to e.$

zhw.
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  • I could do that with taylor or with l'hopital. But how does that help " much " here ? – mick Nov 04 '15 at 01:10
  • Suppose you know this. What is the limit of the numerator? What is the limit of the denominator? – zhw. Nov 04 '15 at 01:16
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    Sorry , I made a typo ! – mick Nov 04 '15 at 01:22
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    Well with the new version, I'm not sure "gross" is the right term. Perhaps "perverse"? (Maybe there's a nice way to see it. Sorry, but I have to clean the lint trap now.) – zhw. Nov 04 '15 at 01:40
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Use the fact that $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$. It is going to be quite lengthy so bear with me: The numerator is $$ \left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n- \left(1+\frac{1}{n}-\frac{1}{n^2}\right)^n = \frac{2}{n^2}\sum_{m=0}^{n-1} \left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1-m} \left(1+\frac{1}{n}-\frac{1}{n^2}\right)^m $$ Write the above as $2/n^2 f(n)$. The denominator has two parts $$ \left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n- \left(1+\frac{1}{n}-\frac{1}{n^2+1}\right)^n = \left(\frac{1}{n^2}+\frac{1}{n^2+1}\right)g(n) $$ where $$g(n) = \sum_{m=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1+m}\left(1+\frac{1}{n}-\frac{1}{n^2+1}\right)^m$$ and the second part $$ \left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n- \left(1+\frac{1}{n}-\frac{1}{n^2(n^2+1)}\right)^n = \frac{1}{n^2}\left(1+\frac{1}{n^2+1}\right)k(n) $$ where $$k(n)=\sum_{m=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1+m}\left(1+\frac{1}{n}-\frac{1}{n^2(n^2+1)}\right)^m$$ so in the end you have $$ \lim_{n\to\infty}\frac{\frac{2}{n^2}f(n)}{\frac{2n^2+1}{n^2(n^2+1)}g(n)+\frac{n^2+2}{n^2(n^2+1)}h(n)}= \lim_{n\to\infty}\frac{2(n^2+1)f(n)}{(2n^2+1)g(n)+(n^2+2)h(n)} $$ Now note that as $n\to \infty$ then $f(n)$ becomes approximately $$f(n) = \sum_{m=0}^{n-1} \left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1-m} \left(1+\frac{1}{n}-\frac{1}{n^2}\right)^m\approx n\left(1+\frac{1}{n}\right)^n\approx ne$$ where $\approx$ means to order of $o(1)$. Actually $g(n)\approx ne\approx h(n)$ too. So the limit reduces to $$\to \lim_{n\to\infty}\frac{2(n^2+1)}{(2n^2+1)+(n^2+2)}=\frac{2}{3}$$ Hopefully I haven't made a mistake anywhere, but at least I think the method is clear.

Hamed
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  • I don't immediately see why the errors in your computation of $f(n)$ are only $o(1/n)$, but you only need them to be $o(n)$, which is easy to see because each of the $n$ terms is converging to $e$ and this convergence is uniform across all the terms. – Eric Wofsey Nov 04 '15 at 01:55
  • You're completely right. Actually I was just thinking about that myself. The order is definitely not $o(1/n)$, at most it is $o(1)$ (I'm not sure about this either). But as you just said $o(n)$ is enough. – Hamed Nov 04 '15 at 02:01
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Perhaps a way to 'see the answer' is to use the MVT: for the numerator, write $f( x) = x^n$, $a = 1 + 1/n -1/n^2$, $b = 1 + 1/n + 1/n^2$. Then $$ f( b) - f(a) = f'(c)(b-a),$$ for some $c \in (a,b)$. Therefore the numerator is
$$ n\left(1 + 1/n + o(1/n^2)\right)^{n-1} 2/n^2.$$ Similarly, in the denominator, using the same trick twice, one gets $$ n\left(1 +1/n + o(1/n^2)\right)^{n-1} ( 2/n^2 + 1/(n^2+1) + 1/n^2(n^2+1) ).$$ Taking the limit (of the ratio!), one gets Hamed's answer of $2/3$.

peter a g
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1

Hint: (for the first version, before multiplying by 2) $$ \left(1+\frac1n\right)^n\leq\left(1+\frac1n+\frac1{n^2}\right)^n\leq\left(1+\frac{1+\epsilon}n\right)^n $$ for large $n$. Hence all partial limits are $e$ and the final one — 0.

Kola B.
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  • Not trying to be rude but : so what ? Far from a solution ? – mick Nov 04 '15 at 01:09
  • @mick The limit of the second term is between $e$ and $e^{1+\epsilon}$, hence is equal to $e$. You can show in the similar way, that the other limits of expressions of the form $(...)^n$ are also $e$. – Kola B. Nov 04 '15 at 01:12
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    Yes. But I made a typo. Sorry. – mick Nov 04 '15 at 01:24
  • Perhaps better to use $$(1+1/n+c/n^2)^n=\exp(n·\ln(1+1/n+c/n^2))\=\exp(n·((1/n+c/n^2)-(1/n+c/n^2)^2/2+O(1/n^3)))\=e·(1+(c-1/2)/n+O(n^2))$$ to get a better quantification of the error. – Lutz Lehmann Nov 04 '15 at 10:50
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You might get some easy simplifications by extending the fraction with the factor $(1-\frac1n)^n$ so that all terms transform to the form $$ \left(1+\frac{c_n}{n^2}\right)^n=1+\frac{c_n}{n}+O(n^{-2}). $$ to obtain \begin{align} &\ \lim_{n \to \infty} \frac{\left(1 - \frac{1}{n^3}\right)^n - \left(1 - \frac{2}{n^2} + \frac{1}{n^3}\right)^n }{ 2 \left(1 - \frac{1}{n^3}\right)^n - \left(1 - \frac{2n^2+1-n}{n^2 (n^2 + 1)}\right)^n - \left(1 - \frac{n^2+2}{n^2 (n^2+1)} + \frac{1}{n^3 (n^2 +1)}\right)^n } \\ \\ &=\lim_{n \to \infty} \frac{\left(1 + O(n^{-2})\right) - \left(1-\frac2n + O(n^{-2}) \right) }{ 2 \left(1 + O(n^{-2})\right) - \left(1 - \frac{2}{n} + O(n^{-2})\right) - \left(1 - \frac{1}{n} + O(n^{-2})\right) } \\ \\ &=\frac 23 \end{align}

Lutz Lehmann
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$$ n\log\left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)=1+\frac xn-\frac1{2n}+O\left(\frac1{n^2}\right) $$ Therefore, $$ \left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)^n=e\left(1+\frac xn-\frac1{2n}\right)+O\left(\frac1{n^2}\right) $$ Thus $$ \begin{align} &\frac{\overbrace{\left(1+\frac1n+\frac1{n^2}\right)^n}^{x=1}-\overbrace{\left(1+\frac1n-\frac1{n^2}\right)^n}^{x=-1}}{2\underbrace{\left(1+\frac1n+\frac1{n^2}\right)^n}_{x=1}-\underbrace{\left(1+\frac1n-\frac1{n^2+1}\right)^n}_{x=-1}-\underbrace{\left(1+\frac1n-\frac1{n^2(n^2+1)}\right)^n}_{x=0}}\\ &=\frac{e\left(1+\frac1n-\frac1{2n}\right)-e\left(1-\frac1n-\frac1{2n}\right)+O\left(\frac1{n^2}\right)}{2e\left(1+\frac1n-\frac1{2n}\right)-e\left(1-\frac1n-\frac1{2n}\right)-e\left(1-\frac1{2n}\right)+O\left(\frac1{n^2}\right)}\\[6pt] &=\frac{\frac2n+O\left(\frac1{n^2}\right)}{\frac3n+O\left(\frac1{n^2}\right)} \end{align} $$

robjohn
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