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.
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 $e^{x}-1-2x &lt;0$ in this interval. Can you take over from here?
We have that
$$\sqrt[n]e=e^{\frac1n}= 1+\frac1n+O\left(\frac1{n^2}\right)$$
then
$$\frac{\sqrt[n]e-1}{n} = \frac1{n^2}+O\left(\frac1{n^3}\right)$$
and we can conclude by limit comparison test.