Estimation of $I_n=\int_1^\infty \frac{1}{1+x+\ldots+x^n} dx$

Question

Let $I_n=\int_1^\infty \frac{1}{1+x+\ldots+x^n} dx$. I'm being asked to determine in 3 different ways whether $\sum_{n\geq 2} I_n$ converges.

I've noticed that $I_n$ rewrites alternatively as $$\int_1^\infty \frac{x-1}{x^{n+1}-1} dx =\int_0^1 \frac{1-x}{1-x^{n+1}} x^{n-2} dx.$$

I've also noticed that $0\leq I_n\leq \int_1^{\infty} \frac 1{x^n} dx = \frac 1{n-1}$, hence the estimate $I_n = O(\frac 1n)$. I'm not sure whether this estimate is tight, or whether it can be refined.

I tried to refine the upper bound as $I_n\leq \int_1^{\infty} \frac 1{x^{n-1} + x^n} dx$, but this looks like a deadend.

Answer 1

Set $k = \lfloor n/2 \rfloor$. For $n \ge 3$ and $x \ge 1$ is $$ 1 + x + \cdots + x^n \ge x^{k+1} + \cdots + x^{2k} \ge k x^{k+1} $$ and therefore $$ I_n \le \int_1^\infty \frac{1}{kx^{k+1}} dx = \frac{1}{k^2} \le \frac{1}{(n/2-1)^2}= \frac{4}{(n-2)^2} \, . $$ It follows that $\sum I_n$ is convergent.