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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?

Rodrigo de Azevedo
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bianco
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3 Answers3

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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 $$\int_0^1\frac{\mathrm dx}{1+x}=\ln(1+x)\Big\vert_0^1.$$

Bernard
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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 $$\lim_{n\to\infty}I=\lim_{n\to\infty}(H_{2n}-H_{n})=\ln 2.$$

stokes-line
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    Perfect, I found that last form to! Could you prove that it tends to $\ln 2$ – bianco Mar 23 '20 at 11:12
  • @mattos: if $\sim$ means here asymptotic equivalence, it is not compatible with addition nor subtraction. – Bernard Mar 23 '20 at 15:36
  • @Bernard It does mean asymptotic equivalence, though I was using it as a rough shorthand for $\lim_{n \to \infty} (H_{2n} - H_{n}) = \lim_{n \to \infty} (\ln 2n - \ln n) = \lim_{n \to \infty} \ln (2n/n) = \dots$. Deleted now. – Matthew Cassell Mar 24 '20 at 01:06
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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 $$\lim_{n\to\infty}\int_0^1\frac{x^n-x^{2n}}{1-x}\ dx=\lim_{n\to\infty}(H_{2n}-H_n)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}=\ln2$$

Ali Shadhar
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