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 ?
 A: 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.
A: 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 $\sum_{n=1}^{\infty} u_n$.
A: Using Euler's Beta function, we are simply asked about the convergence of
$$ \sum_{n\geq 1}\left(1-\frac{\Gamma\left(1+\tfrac{1}{n}\right)^2}{\Gamma\left(1+\tfrac{2}{n}\right)}\right) $$
where $\Gamma(1+x)$ is a regular function in a neighbourhood of $x=0$. Since
$$ \log\Gamma(1+x) = -\gamma x+\frac{\zeta(2)}{2!}x^2-\frac{\zeta(3)}{3}x^3+\frac{\zeta(4)}{4}x^4-\ldots $$
we have that 
$$ 1-\frac{\Gamma\left(1+\tfrac{1}{n}\right)^2}{\Gamma\left(1+\tfrac{2}{n}\right)} \sim \frac{\pi^2}{6n^2} $$
so the given series is convergent for sure. Geometrically, we are saying that if $A_n$ is the area of the region in the $[0,1]\times[0,1]$ square given by $x^{n+1}+y^{n+1}\leq 1 \leq x^n+y^n$, then
$$ A_1+2A_2+3 A_3+\ldots $$
is a convergent series. That can be proved by elementary means, too.
