Does the series $\sum_{n\ge1}\int_0^1\left(1-(1-t^n)^{1/n}\right)\,dt$ converge? 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 ?

Here is a rather elementary solution.
Notice that $u_n$ is the area of the region $D$ in the unit square $[0,1]^2$ defined by $x^n + y^n \geq 1$. Now let $a_n = 2^{-1/n}$ and we split $D$ into three parts,


*

*$ D_1 = \{(x, y) \in [0,1]^2 : x^n + y^n \geq 1 \text{ and } x \leq a_n \} $

*$ D_2 = \{(x, y) \in [0,1]^2 : x^n + y^n \geq 1 \text{ and } y \leq a_n \} $

*$ D_3 = \{(x, y) \in [0,1]^2 : x, y \geq a_n \} $


$\hspace{8em}$ 
Then it is easy to check that $D = D_1 \cup D_2 \cup D_3$ and they are non-overlapping. Also, exploiting the symmetry, we check that $D_1$ and $D_2$ have the same area. So it follows that
\begin{align*}
u_n
&= 2\text{[Area of $D_1$]} + \text{[Area of $D_3$]} \\
&= 2\int_{0}^{a_n} (1 - (1-x^n)^{1/n}) \, dx + (1 - a_n)^2.
\end{align*}
Since $a_n = 1 - \mathcal{O}(\frac{1}{n})$, the term $(1-a_n)^2$ is good. For the integral term, notice that for $x \in [0, a_n]$,
\begin{align*}
1 - (1-x^n)^{1/n}
&= 1 - e^{\frac{1}{n}\log(1-x^n)}
 \stackrel{\text{(1)}}{\leq} -\frac{1}{n}\log(1-x^n) \\
&= \int_{0}^{x} \frac{t^{n-1}}{1-t^n} \, dt
 \stackrel{\text{(2)}}{\leq} \int_{0}^{x} 2t^{n-1} \, dt \\
&= \frac{2}{n}x^n.
\end{align*}
For $\text{(1)}$, we used the inequality $e^t \geq 1 + t$ which holds for all real $t$. The second inequality $\text{(2)}$ follows from the fact that $1-t^n \geq \frac{1}{2}$ for $t \in [0,a_n]$. Thus
$$ u_n \leq \frac{4}{n}\int_{0}^{a_n} x^n \, dx + (1-a_n)^2 \leq \frac{4}{n(n+1)} + (1-a_n)^2 = \mathcal{O}\left(\frac{1}{n^2}\right). $$
This proves the convergence of $\sum_{n=1}^{\infty} u_n$.
