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

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.

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 \} $

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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 $\sum_{n=1}^{\infty} u_n$.

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.