Does $\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^{n^2}\left(\frac{1}{e}\right)^n$ converge?

Question

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

• With the Root Test I get: $e\times \frac{1}{e}=1$, which doesn't determine whether it converges or not.

• With the Ratio Test I have to compute:

$$\lim_{n\to \infty}\left|\frac{\left(1+\frac{1}{n+1}\right)^{{(n+1)}^2}\left(\frac{1}{e}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^{n^2}\left(\frac{1}{e}\right)^n}\right|=\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{2n+1}\left(\frac{1}{e}\right)$$ \begin{align} \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{2n+1}&=\exp\lim_{n\to \infty}\left(1+\frac{1}{n}-1\right)(2n+1)\\ &=\exp\lim_{n\to\infty}2+\frac{1}{n}\\ &=\exp2=e^2 \end{align}

$e^2\times \frac{1}{e}=e>1$, so it diverges

Is this correct? Im kinda new to this concept so I need a little help.

Answer 1

You limit computations are not exactly correct even though they give the final good answer. However to exactly compute the limit:

\begin{align} \lim_{n\to \infty} \left(1 + \frac1n\right)^{2n+1} &= \lim_{n\to \infty} \left(\left(1 + \frac1n\right)^n\right)^{2}\left(1+\frac1n\right) = e^2\times 1 = e^2 \end{align}

Answer 2

A necessary condition for convergence is that the $n^{th}$ term should go to zero. Let $\lim_{n \to \infty}{(1+1/n)}^{n^2}{(1/e)}^{n}=A.$ Then

$\implies \lim_{n \to \infty}{n^2}\log{(1+1/n)}+{n}\log{1/e}=\log A $

$\implies \lim_{n \to \infty}{n^2}{(1/n-1/{2n^2}+...)}-{n}=\log A $

$\implies \log A=-1/2$

Contradiction, hence It is divergent. What you have done is also correct apart from what @youem mentioned.

Answer 3

You don't need either of these tests, because a necessary condition for convergence is that the general term tends to $0$.

Now, using Taylor-Young's expansion, we have \begin{align} \biggl(1+\frac1n\biggr)^{n^2}&=\mathrm e^{n^2\log\bigl(1+\frac1n\bigr)}=\mathrm e^{n^2\log\bigl(1+\frac1n\bigr)}\\ &=\mathrm e^{n^2\bigl(\frac1n-\frac1{2n^2}+o\bigl(\frac 1{n^2}\bigr)\bigr)}=\mathrm e^{n-\frac1{2}+o\bigl(1\bigr)} \end{align} so that $$\biggl(1+\frac1n\biggr)^{n^2}\frac1{\mathrm e^n}= \mathrm e^{-\frac1{2}+o\bigl(1\bigr)},$$ which tends to $\dfrac1{\sqrt{\mathrm e}}$.

Answer 4

As an alternative, using that $\log (1+x)\ge x-\frac {x^2}2$ we obtain

$$\left(1+\frac{1}{n}\right)^{n^2}=e^{n^2\log\left(1+\frac{1}{n}\right)}\ge e^{n-\frac12}=\frac{e^n}{\sqrt e}$$

and therefore

$$\left(1+\frac{1}{n}\right)^{n^2}\left(\frac{1}{e}\right)^n\ge \frac{e^n}{\sqrt e}\cdot \frac1{e^n}=\frac{1}{\sqrt e}$$

Refer also to the related

• Mean value theorem and inequality proving