Absolute convergence and conditional convergence of the sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot(1+\frac{1}{n})^n}{n}$

Question

I am trying to solve an exercise from an old exam from my university which no solution is uploaded.

Basically, I have to determine if the series:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot(1+\frac{1}{n})^n}{n}$$

is convergent and also check if it is absolutely convergent.

To know if it is convergent I am trying to apply the Leibnitz Criteria for alternating series, that states that:

$\sum_{n=1}^{\infty} (-1)^{n}a_n$ convergent iff $a_n$ is decreasing and its limit is $0$

For this series, if we call $a_n = \frac{(1+\frac{1}{n})^n}{n}$ I have proven that it converges to $0$ but I do not see an easy way to prove that It is decreasing.

For the absolute convergence I have tried several criteria but none of them works for me.

Thank you in advance.

Answer 1

Yes, the sequence $a_n=\frac{\left(1+\frac1n\right)^n}{n}$ is decreasing and the proof is fairly simple: $$\frac{a_{n+1}}{a_n} =\frac{(\frac{n+2}{n+1})^{n+1}}{n+1}\cdot\frac{n}{(\frac{n+1}{n})^n} =\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}<1$$ where at the last step we just note that $n(n+2)=(n+1)^2-1<(n+1)^2$. Since you already know that $a_n\to 0$, by Leibniz criterion, the series is convergent.

The series is NOT absolute convergent by limit comparison test: $$\left|\frac{(-1)^{n+1}\cdot(1+\frac{1}{n})^n}{n}\right|\sim \frac{e}{n}$$ and $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent.

Answer 2

This Legendary inequality might help you: $$\frac{2}{n}\le \frac{\left(1+\frac1n\right)^n}{n}\le\frac en.$$ Which shows that $\displaystyle\lim_{n\to\infty}\frac{\left(1+\frac1n\right)^n}{n}=0.$

Edit: Apart from that, let $a_n=\frac{\left(1+\frac1n\right)^n}{n}$ then by Limit comparison test using the series $\sum_n\frac{1}{n}=\sum_n b_n$ we have $$\lim_{n\to\infty} \frac{a_n}{b_n}=e\text{, a non zero real number.}$$ Which means $\sum_n a_n$ and $\sum_n b_n$ has same convergence tendency (i.e. $\sum a_n$ converges (diverges) $\iff \sum b_n$ converges (diverges)). Now $\sum_n b_n$ is a Harmonic series which means it is divergent and so we have our resyult.

Answer 3

Let $\log x$ denote the natural log function, then when $|z|<1$ there is

$$ \log(1+z)=z-\frac z2+{z^3\over3}-{z^4\over4}+\cdots $$

This suggests that when $n$ is large

$$ n\log\left(1+\frac1n\right)=1+O\left(\frac1n\right) $$

This means if we define

$$ a_n=e-\left(1+\frac1n\right)^n $$

then $a_n=O(1/n)$ when $n$ grows large. As a result, we have

$$ \sum_{n\le N}{(-1)^{n+1}\over n}\left(1+\frac1n\right)^n=e\sum_{n\le N}{(-1)^{n+1}\over n}+\sum_{n\le N}{(-1)^{n+1}a_n\over n} $$

The first sum on the right converges due to Leibniz's test, and the second sum converges absolutely because the summand is $O(1/n^2)$ when $n$ goes to infinity.