Does $\sum_{n=1}^\infty \frac{\sqrt[n]e-1}{n}$ converge? Does $$\sum_{n=1}^\infty \frac{\sqrt[n]e-1}{n}$$ converge or diverge?
Could you please help me out with this one? I tried the ratio test, but it's inconclusive...
I also tried to use the inequality $e^x \ge x+1$, but it doesn't get me anywhere.
 A: The series is convergent. $e^{x}-1-2x$ is decreasing in $(0, \ln 2)$ since its derivative is negative here. Since this function vanishes at $0$ it follows that $e^{x}-1-2x <0$ in this interval. Can you take over from here?
A: You have
$$e^{1/n}-1 \sim \frac{1}{n}$$
so
$$\frac{\sqrt[n]{e}-1}{n} \sim \frac{1}{n^2}$$
so the series is convergent by comparison.
A: You can apply the direct comparision test if you construct a suitable series which forms an upper bound.
Let be $n>1$, then by definition we know that $e^{x}:=\sum\limits_{j=0}^{\infty}\frac{x^j}{j!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\cdots$.
Hence, $$e^{\frac{1}{n}}-1=\lim\limits_{k\to\infty}\sum\limits_{j=0}^{k}\frac{\left({\frac{1}{n}}\right)^j}{j!}-1=\sum\limits_{j=1}^{\infty}\frac{\left({\frac{1}{n}}\right)^j}{j!}.$$ As
$
\Big|\frac{\left({\frac{1}{n}}\right)^j}{j!}\Big|\leq \left({\frac{1}{n}}\right)^j$ for all $j$, we can conclude that
$$
\sum\limits_{j=1}^{k}\frac{\left({\frac{1}{n}}\right)^j}{j!}\leq \sum\limits_{j=1}^{k}\left({\frac{1}{n}}\right)^j=\frac{n}{n-1}\left(\frac{1}{n}-\left(\frac{1}{n}\right)^{k+1}\right).
$$
Hence, $e^{\frac{1}{n}}-1=\lim\limits_{k\to\infty}\sum\limits_{j=1}^{k}\frac{\left({\frac{1}{n}}\right)^j}{j!}\leq \frac{1}{n-1}$. This yields the following bound for each summand of $\sum\limits_{n=2}^{k}\frac{e^{\frac{1}{n}}-1}{n}$:
$$
\Big|\frac{e^{\frac{1}{n}}-1}{n}\Big|\leq \frac{1}{n(n-1)}.
$$
We know that the series $\left(2+\sum\limits_{n=2}^{\infty}\frac{1}{n(n-1)}\right)$ converges because $\sum\limits_{n=2}^{k}\frac{1}{n(n-1)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}\cdots -\frac{1}{k}=1-\frac{1}{k}$ and therefore $\sum\limits_{n=2}^{\infty}\frac{1}{n(n-1)}=1$. So from the direct comparision test it follows that $\sum\limits_{n=1}^{\infty}\frac{e^{\frac{1}{n}}-1}{n}$ converges (absolutely).
