# Limit calculation known formula

can anyone explain to me why this limit is true? I've tried to solve it using the known limit, but I got 1

$$\lim_{n \rightarrow \infty}\frac{(1+\frac1n)^{n^2}}{e^n} = \frac{1}{\sqrt{e}}.$$

• Sorry for the poor question and title formulation, but the quality standards prohibited me from making it simple. – CodeHoarder Jan 3 '17 at 14:44
• oh the question is clear – user394255 Jan 3 '17 at 15:00
• I hope you do know that $$\lim_{n \rightarrow \infty} \left( 1+\frac{1}{n} \right)^n = e .$$ – user399481 Jan 3 '17 at 15:08
• Alright and e^n/e^n =1 – CodeHoarder Jan 3 '17 at 15:10
• And, now your question is unclear again. What are you saying by "but I got 1" ? – user399481 Jan 3 '17 at 15:11

## 2 Answers

You could take the logarithm and rewrite: $$\ln \frac{\left(1+\frac1n \right)^{n^2}}{e^n} = n^2 \ln \left(1+\frac1n \right)-n =\frac{\ln \left(1+\frac1n \right)-\frac1n}{\frac{1}{n^2}}$$ For $n \to \infty$, you get the indeterminate form $\tfrac{0}{0}$ so you can apply l'Hôpital's rule: $$\lim_{n \to \infty} \frac{\ln \left(1+\frac1n \right)-\frac1n}{\frac{1}{n^2}}=\lim_{n \to \infty} \frac{\frac{-\frac{1}{n^2}}{\left(1+\frac1n \right)}+\frac{1}{n^2}}{\frac{-2}{n^3}} =\cdots=\lim_{n \to \infty} \frac{n}{-2(n+1)}=-\frac{1}{2}$$ So with $\lim_{n \to \infty} u_n = e^{\lim_{n \to \infty} \ln u_n}$, you have: $$\lim_{n \to \infty}\frac{\left(1+\frac1n \right)^{n^2}}{e^n} = e^{-\tfrac{1}{2}}=\frac{1}{\sqrt{e}}$$

Let $a_n$ be the sequence of your problem. Then:

$\ln a_n=n(n\ln(1+\frac{1}{n})-1)$

By the Taylor series of the logarithm, we have that there exists $x_n \in (0.\frac{1}{n})$:

$\ln(1+\frac{1}{n})=\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3(x_n+1)^3} \Rightarrow n(n\ln(1+\frac{1}{n})-1)=-\frac{1}{2}+\frac{1}{3n(x_n +1)^3}$

Since $x_n\rightarrow 0$ when $n\rightarrow +\infty$, we take the limits in the previous equality and conclude that $\ln a_n\rightarrow -\frac{1}{2}\Rightarrow a_n\rightarrow e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}$