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I have to find the sum of: $$1+\frac13-\frac12-\frac14-\frac16+\frac15+\frac{1}{7}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}+\ldots$$

My attempt:

$$\left( 1+\frac13-\frac12-\frac14-\frac16\right)+\left(\frac15+\frac{1}{7}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}\right)+\ldots$$

\begin{eqnarray*}\sum_{n=0}^{+\infty}\left(\frac{1}{4n+1}+\frac{1}{4n+3}-\frac{1}{6n+2}-\frac{1}{6n+4}-\frac{1}{6n+6}\right)&=&\int_{0}^{1}\frac{1+x^2}{1-x^4}-\frac{x+x^3+x^5}{1-x^6}\,dx\\&=&\log2\end{eqnarray*}

But the answer is given as $\frac12\log{\frac83}$.Where did I make a mistake?

Also,could I get the solution using $\log 2=1-\frac12+\frac13-\frac14+\frac15-\frac16+\ldots$?If so,how?

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    $\begingroup$ How did you convert series to integral? Thanks $\endgroup$ – SJ. Dec 30 '17 at 7:34
  • $\begingroup$ @samjoe I used $\int_{0}^{1}x^m\,dx = \frac{1}{m+1}$ $\endgroup$ – PiGamma Dec 30 '17 at 7:34
  • $\begingroup$ Are you supposed to use generalized harmonic numbers ? $\endgroup$ – Claude Leibovici Dec 30 '17 at 9:03
  • $\begingroup$ @ClaudeLeibovici I have edited the question $\endgroup$ – PiGamma Dec 30 '17 at 9:09
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    $\begingroup$ For more about this, see questions 2576166, 2575967, 2554102, 911293. $\endgroup$ – cgiovanardi Dec 30 '17 at 11:56
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The series is not absolutely convergent; therefore, you cannot move around terms freely.In other words,

\begin{eqnarray*}\sum_{n=0}^{+\infty}\left(\frac{1}{4n+1}+\frac{1}{4n+3}-\frac{1}{6n+2}-\frac{1}{6n+4}-\frac{1}{6n+6}\right)\neq \sum_{n=0}^{+\infty}\left(\frac{1}{4n+1}+\frac{1}{4n+3}\Bigg)-\\-\sum_{n=0}^{\infty}\Bigg(\frac{1}{6n+2}+\frac{1}{6n+4}+\frac{1}{6n+6}\right) \end{eqnarray*}

In fact, both terms on the right hand side of the above equation are both diverging to $\infty$, which should make your mistake even obvious.

If you want a hint, then consider $$f(t) = \sum_{n=0}^{+\infty}\left(\frac{t^{4n+1}}{4n+1}+\frac{t^{4n+3}}{4n+3}-\frac{t^{6n+2}}{6n+2}-\frac{t^{6n+4}}{6n+4}-\frac{t^{6n+6}}{6n+6}\right)$$ on $[0,1)$ and use Abel's continuity theorem and differentiation.

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