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I need to find the limit: $$\lim_{n\to \infty } \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n$$ So I know that the limit is $1$.
Using Squeeze theorem $$? \leq \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n \leq \left(\frac{e}{e}\right)^n \rightarrow\ 1 $$ What should be instead $?$ ? Is it possible to solve in another way?
Unfortunately, I can't use L'Hôpital Rule or Series Expansion in this task.

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Note that:

$$\left(\frac{(1 + \frac{1}{n^2})^{n^2}}{(1 + \frac{1}{n^2})^{n^2+1}}\right)^n=\left(1 + \frac{1}{n^2}\right)^{-n}\leq \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n \leq \left(\frac{e}{e}\right)^n=1$$

Since:

$$\left(1 + \frac{1}{n^2}\right)^{-n}\to 1$$

By Squeeze Theorem:

$$\lim_{n\to \infty } \left(\frac{(1 + \frac{1}{n^2})^{n^2}}{e}\right)^n=1$$

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    $\begingroup$ The use of $e\leq (1+1/n)^{n+1}$ with $n$ replaced by $n^2$ was a smart move. +1 $\endgroup$ – Paramanand Singh Dec 18 '17 at 14:54
  • $\begingroup$ @ParamanandSingh Thanks that's a very useful upper bounding term for e. You appreciation is an honor to me! $\endgroup$ – user Dec 18 '17 at 14:57
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If $$A=\lim_{n\to \infty } \left(\dfrac{\left(1 + \dfrac1{n^2}\right)^{n^2}}e\right)^n$$

$$\ln A=\lim_{n\to \infty }n^3\ln\left(1+\dfrac1{n^2}\right)-n$$

Set $1/n=h$ and

use Are all limits solvable without L'Hôpital Rule or Series Expansion

or L'Hôpital Rule

or Series Expansion

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Note that for $x>0$ the logarithm function satisfies the fundamental inequality (prove it yourself, it is easy) $$\frac{x^{2}}{1+x^{2}}\leq \log(1+x^{2})\leq x^{2}$$ or $$\frac{-x}{1+x^2}\leq \frac{\log(1+x^2)}{x^{3}}-\frac{1}{x}\leq 0$$ Putting $x=1/n$ we get $$-\frac{n} {1+n^2}\leq n^3\log(1+n^{-2})-n\leq 0$$ And using Squeeze theorem we have $n^3\log(1+n^{-2})-n\to 0$. On exponentiating we see that the limit in question is $1$.

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