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

Question

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.

Possible duplicate of Calculate the following: $\lim \limits_{n \to \infty} \sqrt[n]{e^n+(1+\frac{1}{n})^{n^2}}$ — , Dec 18 '17 at 13:28
Your brackets are off, it's hard to interpret what the location of the $n^2$ is. — Matti P., Dec 18 '17 at 13:28

score 4 · Answer 1 · answered Dec 18 '17 at 13:31

4

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

answered Dec 18 '17 at 13:31

lab bhattacharjee

274,582

Unfortunately, I can't use L'Hôpital Rule or Series Expansion in this task – A. Miller Dec 18 '17 at 13:49
1

@A.Rokinsky, How about using the idea of $L3$ here https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion in the best answer – lab bhattacharjee Dec 18 '17 at 13:51

Paramanand Singh · Answer 2 · 2017-12-18T17:26:20.440

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 $1$.

user · Accepted Answer · 2017-12-18T14:33:28.493

3

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:

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

edited Dec 18 '17 at 14:33

answered Dec 18 '17 at 13:57

user

154,566

1

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

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

3 Answers3