I was willing to determine the sum of following $$S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$$

I tried the following \begin{align*} S=&\sum\limits_{n=0}^\infty (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right)\\ =&\sum\limits_{n=0}^\infty (-1)^n\left(\int_0^1 x^{2n-2}dx+\int_0^1 x^{2n-1} dx\right)\\ =&\int_0^1 \sum\limits_{n=0}^\infty (-1)^n\left(x^{2n-2}+x^{2n-1}\right)\\ =&\int_0^1 [x^{-2} \sum\limits_{n=0}^\infty (-1)^nx^{2n}+x^{-1} \sum\limits_{n=0}^\infty (-1)^nx^{2n}] \end{align*} and don't know after his what to do . Can you please help me on this regard?

Thanking you in advance

  • 5
    $\begingroup$ you have two geometric series. sum them up and take the limit $x\rightarrow1^-$ $\endgroup$ – tired May 18 '15 at 8:36
  • 4
    $\begingroup$ I suppose that the summations start at $n=1$ instead of $n=0$ $\endgroup$ – Claude Leibovici May 18 '15 at 8:41

Note that $$S=A+B$$ where $\displaystyle A=\sum_{k\ge 1}\frac{(-1)^{k-1}}{2k-1},\ B=\sum_{k\ge 1}\frac{(-1)^{k-1}}{2k}$. Clearly, $B=\displaystyle \frac{\ln 2}{2}$. and, $A=\tan^{-1}1=\pi/4$ which you can find out using your technique as below $$A=\sum_{k\ge 1}(-1)^{k-1}\int_0^1 x^{2k-2}dx\\=\int_{0}^1\sum_{k\ge 1}(-1)^{k-1}x^{2k-2}dx\quad(\mbox{Use Fubini to justify the change of order})\\=\int_{0}^1 \frac{dx}{1+x^2}=\tan^{-1}1=\pi/4\\ \mbox{similarly, }\ B=\int_{0}^1 \sum_{k\ge 1}(-1)^{k-1}x^{2k-1}dx\\=\int_0^1 \frac{x}{1+x^2}dx\\=\frac{\ln 2}{2}$$

  • $\begingroup$ Well you can use DCT here with equal ease but Fubini's theorem also work, with the measures in mind as the counting measure on $\mathbb{N}$ and the Lebesgue measure on $[0,1]$. Please see this for more clarification. $\endgroup$ – Samrat Mukhopadhyay Dec 21 '15 at 5:51

Couldn't you just write this as

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} + \frac12 \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+1} = \frac{\pi}{4} + \frac12 \log{2}$$

  • $\begingroup$ But that would be too easy. +1 :-)) -Mark $\endgroup$ – Mark Viola Dec 18 '15 at 20:26

Let $$S = \sum_{n = 1}^{\infty} (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right).$$ Then

$\begin {eqnarray} S & = & \sum_{k = 1}^{\infty} (-1)^{2k}\left(\frac{1}{4k-1}+\frac{1}{4k}\right) + \sum_{k = 1}^{\infty} (-1)^{2k+1}\left(\frac{1}{4k+1}+\frac{1}{4k+2}\right) \nonumber \\ & = & \sum_{k = 1}^{\infty} \left(\frac {1} {4k - 1} - \frac {1} {4k + 1}\right) + \sum_{k =1}^{\infty} \left(\frac {1} {4k} - \frac {1} {4k + 2}\right), \end {eqnarray}$

which is easier to proceed now.

  • $\begingroup$ How is your final result easier?? $\endgroup$ – Mark Viola Dec 18 '15 at 20:24
  • $\begingroup$ Because it's not a strangely alternate series anymore. $\endgroup$ – user98186 Dec 18 '15 at 20:33
  • $\begingroup$ But how is that easier? How would you propose that one proceed? Use Digamma function IDs? $\endgroup$ – Mark Viola Dec 18 '15 at 21:04

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