Limit of $\int_0^1 \frac{x^n-x^{2n}}{1-x}\text{d}x$ I have to find the value of  $$\lim_{n\to\infty} \int_0^1 \frac{x^n-x^{2n}}{1-x}\text{d}x$$ with $n\in\mathbb{N}$. The result is $\ln 2$ but i find no way to prove it. I thought that $$H_n=\int_0^1 \frac{1-x^n}{1-x}\text{d}x$$ would have helped but I came up with nothing. The fact that the result is the the same as the sum of alternating harmonic series made me think i should find a way to make a $(-1)^k$ pop out from somewhere but all my attempts failed… Any idea?
 A: Your integral is $H_{2n}-H_n$ which using known behavior of $H_n$ at large $n$, that is $H_{n\gg 1} = \ln n + C$, gives $$\lim_{n\to\infty}I=\lim_{n\to\infty}(H_{2n}-H_{n})=\ln 2.$$ 
A: Just with Riemann sums:
First rewrite the integrand as
$$ \frac{x^n-x^{2n}}{1-x}=x^n\frac{1-x^{n}}{1-x}=x^n(1+x+\dots+x^{n-1}) 
=\sum_{k=0}^{n-1}x^{n+k}, $$
whence the value of the integral
$$\int_0^1\frac{x^n-x^{2n}}{1-x}\,\mathrm dx=\sum_{k=1}^{n}\frac1{n+k}
=\frac1n\sum_{k=1}^{n}\frac1{1+\cfrac kn}.$$
Now the latter expression is the lower Riemann sum for the integral
$$\int_0^1\frac{\mathrm dx}{1+x}=\ln(1+x)\Big\vert_0^1.$$
A: We have 
$$\overline{H}_{2n}=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2n}(-1)^{k-1}\int_0^1 x^{k-1}\ dx=\int_0^1\sum_{k=1}^{2n}(-x)^{k-1}\ dx$$
$$=\int_0^1\frac{1-x^{2n}}{1+x}\ dx=\ln2-\int_0^1\frac{x^{2n}}{1+x}\ dx=H_{2n}-H_n$$
where the last result follows from using $\displaystyle\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$
so $$\lim_{n\to\infty}\int_0^1\frac{x^n-x^{2n}}{1-x}\ dx=\lim_{n\to\infty}(H_{2n}-H_n)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}=\ln2$$
