# 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 ?

• I understand that you know if the series converges or not. Can you tell us? If we know what we should prove, it would be easier to do. (For example, if I know that it converges, I will try to find an upper bound). Commented Jan 14, 2017 at 15:05
• it is easy to show that the second part of the integral equals $\frac{\Gamma(1+1/n)^2}{\Gamma(1+2/n)}$ which is asymptotic to $1-\frac{\pi^2}{6n^2}$ for $n\rightarrow \infty$. This means $u_n\sim \frac{\pi^2}{6n^2}$ and the series converges.. Commented Jan 14, 2017 at 15:09
• @ajotaxe Haha ! That's right. Well, if my proof is correct, it converges. Commented Jan 14, 2017 at 15:09
• @tired Looks like this calculation wasn't easy (nor natural) for me. I don't see at first glimpse from where this result comes from, but I will try to find ... Commented Jan 14, 2017 at 15:12
• this is more a less a consequence of the integral rep of the beta function en.wikipedia.org/wiki/Beta_function and known properties of the gamma function Commented Jan 14, 2017 at 15:14

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.

• Nice, I should have speculated that this inequality was enough for the proof. (+1) Commented Jan 14, 2017 at 19:17
• @sangchul Thank you! Much appreciative. -Mark Commented Jan 14, 2017 at 19:27
• Great proof and sharp estimation, as we know with the equivalent given by Jack d'Aurizio. Thank you. Commented Jan 14, 2017 at 21:20
• Thank you. I had already seen the comment left by @tired and the subsequent "write-up" by Jack. So, I pursued a different approach. Note that we can obtain lower bound by noting (Bernoulli's Inequality) $(1-t^n)^{1/n}\le 1-\frac1n t^n$. The result is $$1-\int_0^1 (1-t^n)^{1/n}\,dt\ge \frac{1}{n(n+1)}$$ Commented Jan 14, 2017 at 21:25

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

• This is precisely the solution I had in mind. Thank you for your response. Commented Jan 14, 2017 at 19:03
• @Adren Dr. MV gave a very nice answer which greatly simplifies my computation. Commented Jan 14, 2017 at 19:31
• (+1) The Dirichlet hyperbola method (sort of) is a nice touch. Commented Jan 14, 2017 at 19:45

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.