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How to calculate this series?

$$ \sum_{n=0}^{\infty} (-1)^n \frac{0.5^{n+1}}{(2n+1)(2n+2)}$$

I try to operate: $ \sum_{n=0}^{\infty} (-1)^n \frac{0.5^{n+1}}{(2n+1)(2n+2)}=$$ \sum_{n=0}^{\infty} (-1)^n 0.5^{n+1}(\frac{1}{2n+1}-\frac{1}{2n+2})$ But it does not work

1 Answers1

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$$=\dfrac 1a\sum_{n=0}^\infty\dfrac{a^{2n+1}}{2n+1}-\dfrac 1{a^2}\sum_{n=0}^\infty \dfrac{a^{2n+2}}{2n+2}$$

Now $\ln(1+x)+\ln (1-x)=?$

and $\ln(1+x)-\ln(1-x)=?$

Here $a^2=-.5$

lab bhattacharjee
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  • But how to get the $a^{2n+1}$? I mean this is $a^n$ but not $2n+1$. –  Oct 24 '18 at 07:04
  • @Tinzoe-Yui, The version of the post was $$\sum_{n=0}^{\infty} (-1)^n \frac{0.5^n}{(2n+1)(2n+2)}=\sum_{n=0}^{\infty} \frac{(-0.5)^n}{(2n+1)(2n+2)}$$ I replaced $-.5$ with $$a^2$$ – lab bhattacharjee Oct 24 '18 at 07:07
  • I see. But the answer appears $\arctan(1/2)$, why? If follow your method, it only appears the term $\log(3/2)$. –  Oct 24 '18 at 07:16
  • @Tinzoe-Yui, How? Remember $a=\sqrt{\dfrac{-1}2}=?$ Also, https://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots – lab bhattacharjee Oct 24 '18 at 07:25
  • Thanks! I think the answer uses the Taylor expansion of the $\arctan x$. –  Oct 24 '18 at 08:55