8

Let $$v_n=\dfrac 1 {n+1}\sum_{k=0}^n \dfrac 1 {k+1}$$ We wanna study the sum $$S=\sum_{n=0}^{+\infty}(-1)^n v_n$$

The problem says we should first find $\omega(x)$ s.t. $$v_n=\int_0^1 x^n\omega(x)dx$$ Then we'll have $S=\int_0^1\dfrac {\omega(x)} {1+x}dx$, but I can't find such $\omega(x)$. What's the idea of constructing such integral?

Did
  • 279,727
pqros
  • 830

2 Answers2

10

Recall the geometric series:

$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}$$

Integrate both sides from zero to one,

$$\sum_{k=0}^n\frac1{k+1}=\int_0^1\frac{1-x^{n+1}}{1-x}\ dx$$

It thus follows that

$$v_n=\int_0^1\frac{1-x^{n+1}}{(n+1)(1-x)}\ dx$$

Apply integration by parts to get

$$v_n=-\int_0^1x^n\ln(1-x)\ dx$$

for every $n$, hence $$S=-\int_0^1\frac{\ln(1-x)}{1+x}\ dx$$

Did
  • 279,727
Simply Beautiful Art
  • 74,685
  • 2
    To find $\omega$, apply an integration by parts to your formula for $v_n$, with $$u(x)=\frac{1-x^{n+1}}{n+1}\quad v(x)=-\log(1-x)\quad u'(x)=-x^n\quad v'(x)=\frac1{1-x}$$ hence $$v_n=-\int_0^1x^n\log(1-x)dx$$ that is, $$\omega(x)=-\log(1-x)$$ – Did Mar 01 '17 at 13:03
  • 1
    Thanks @Did, I was wondering how to remove $n$ from my integral. – Simply Beautiful Art Mar 01 '17 at 13:04
  • 2
    Yep. To be integrated to your answer. +1 in advance. – Did Mar 01 '17 at 13:05
5

After we get $$ S = -\int_{0}^{1}\frac{\log(1-x)}{1+x}\,dx = -\int_{0}^{1}\frac{\log x}{2-x}\,dx =-\left.\frac{d}{d\alpha}\int_{0}^{1}\frac{x^{\alpha}\,dx}{2-x}\,\right|_{\alpha=0^+}$$ we also have $$ S = \text{Li}_2\left(\frac{1}{2}\right) = \color{red}{\frac{\pi^2}{12}-\frac{\log(2)^2}{2}}$$ by the dilogarithm reflection formula.

Community
  • 1
Jack D'Aurizio
  • 353,855