TITLE: Does the series $\sum_{n\ge1}\int_0^1\left(1-(1-t^n)^{1/n}\right)\,dt$ converge?
QUESTION [8 upvotes]: Here is a question that I've been working on, a few years ago. I do know how to solve it but I am convinced that it deserves at least another point of view ...
I will post my own solution soon (within a week, at most) and I hope that - meanwhile - other people will suggest various approaches.
Consider, for all $n\in\mathbb{N^\star}$ :
$$u_n=\int_0^1\left(1-(1-t^n)^{1/n}\right)\,dt$$
Does the series $\sum_{n\ge 1}u_n$ converge ?

REPLY [6 votes]: We have 
$$\begin{align}
(1-t^n)^{1/n}&=e^{\frac1n \log(1-t^n)}\\\\
&\ge 1+\frac1n\log(1-t^n)
\end{align}$$
from which 
$$\begin{align}
\int_0^1 (1-t^n)^{1/n}\,dt&\ge 1+\frac1n \int_0^1 \log(1-t^n)\,dt\\\\
&=1+\frac1{n^2} \int_0^1 \frac{\log(1-t)}{t}t^{1/n}\,dt\\\\
&\ge 1+\frac{1}{n^2}\int_0^1 \frac{\log(1-t)}{t}\,dt\\\
&= 1-\frac{\pi^2}{6n^2} 
\end{align}$$
Hence, 
$$0\le \int_0^1 \left(1-(1-t^n)^{1/n}\right)\,dt\le \frac{\pi^2}{6n^2}$$
By the comparison test, the series converges.