how can I find the convergence of the integral $\displaystyle\int_{-1}^{1}\frac{1-x^n}{1-x}$ , $ x \in (-1,1)$ I want to check the convergence of the integral 
$\displaystyle\int_{-1}^{1}\frac{1-x^n}{1-x}$ , for $ x \in (-1,1)$ 
but i don't know what to do. Every theory I know it is not working. 
Can someone help me?
 A: If $n$ is a positive integer, then
$$\frac{1-x^n}{1-x}=1+x+\dots+x^{n-1}$$
A: Well using division you have

$${1-x^n\over 1-x} = \sum_{k=0}^{n-1} x^k$$

Taking absolute values we see
$$\left|{1-x^n\over 1-x}\right|=\left|\sum_{k=0}^{n-1}x^k\right|\le \sum_{k=0}^{n-1}|x|^k\le \sum_{k=0}^{n-1}1 = n$$
But then your integral converges because $$0\le \left|\int_{-1}^1{1-x^n\over 1-x}\,dx\right|\le \int_{-1}^1 n \,dx = 2n<\infty.$$
And absolute convergence implies convergence.
A: $$\int_{-1}^{1}\frac{1-x^n}{1-x}\,dx = \sum_{k=0}^{n-1}\int_{-1}^{1}x^k\,dx \stackrel{(i)}{=}2\sum_{j<\frac{n}{2}}\int_{0}^{1}x^{2j}\,dx=2\sum_{j<\frac{n}{2}}\frac{1}{2j+1}\tag{1}$$
where in $(i)$ we have exploited the fact that the integral of an odd, integrable function over a symmetric interval with respect to the origin is simply zero. Since $x\leq \frac{1}{2}\log\left(\frac{x+1}{x-1}\right)$ for any $x\in[0,1)$,
$$\int_{-1}^{1}\frac{1-x^n}{1-x}\,dx =2+ 2\sum_{1\leq j<\frac{n}{2}}\frac{1}{2j+1}\leq 2+\sum_{1\leq j<\frac{n}{2}}\log\frac{2j+2}{2j}\leq \color{red}{2+\log( n+1)}\tag{2} $$
for any $n\geq 1$.
