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Calculate the summation $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$.

So I said:

Mark $x = \frac{2}{3}$. Therefore our summation is $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{x^{2n-1}}{(2n+1)}}$.

But how do I exactly get rid of the $(-1)^n$? Also I notice it is a summation of the odd powers of $x$, how can I convert it to a full sum? (I know I should subtract from the full sum) but the signs of this summation is different than the signs of the full sum

jimjim
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TheNotMe
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3 Answers3

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One can notice that $$\frac{1}{x^2}\int x^{2n}\mathrm dx=\frac{x^{2n-1}}{(2n+1)}$$ So:

$$\sum_{n=1}^{\infty} {(-1)^n \frac{x^{2n-1}}{(2n+1)}}=\frac{1}{x^2}\sum_{n=1}^{\infty} {(-1)^n }\int x^{2n}\mathrm dx=\frac{1}{x^2}\int \left(\sum_{n=1}^{\infty} {(-1)^n }x^{2n}\right)\mathrm dx$$ And (keeping in mind that $x\leq 1$) $$\sum_{n=1}^{\infty} {(-1)^n }x^{2n}=-\frac{x^2}{x^2+1}$$ Then $$\sum_{n=1}^{\infty} {(-1)^n \frac{x^{2n-1}}{(2n+1)}}=\frac{1}{x^2}\int \left(-\frac{x^2}{x^2+1}\right)\mathrm dx=\frac{\tan^{-1}(x)-x}{x^2}$$

Caran-d'Ache
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    I'm confused about the point following "So:". $x$ is a constant. How can you say that the sum in the first equation is equal to an indefinite integral in $x$ in the second equation? – Jack M Jul 04 '13 at 12:08
  • @JackM Well, it the idea that I wanted to show. Strictly speaking you can put it lik $\frac{1}{x^2}\int_0^x t^{2 n} \mathrm dt$ – Caran-d'Ache Jul 04 '13 at 17:15
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Since $$ \tan^{-1}(x)=\sum_{k=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1} $$ we get $$ \sum_{n=1}^\infty(-1)^n\frac{x^{2n-1}}{2n+1} =\frac{\tan^{-1}(x)-x}{x^2} $$ For $x=\frac23$, the sum is approximately $-0.176994142017973$.

Without knowing the series for $\tan(x)$ first

It is not too difficult to notice that if we multiply the series by $x^2$ and differentiate, we get $$\sum_{k=1}^\infty(-1)^kx^{2k}=\frac1{1+x^2}-1$$ We get that by using the geometric series formula. Now we integrate and divide by $x^2$ to undo what was just done.

robjohn
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  • Let us assume that we don't know that that sum is (almost) equal to $arctan(x).$ Is there any other way of doing it? – TheNotMe Jul 04 '13 at 09:51
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    @TheNotMe: obviously, if you want to evaluate the sum exactly, you are going to have to realize the association to $\tan^{-1}(x)$ at some point. You could notice that if you multiply your series by $x^2$ and differentiate, you get $$\sum_{k=1}^\infty(-1)^kx^{2k}=\frac1{1+x^2}-1$$ We get that by using the geometric series formula. Now integrate and divide by $x^2$ to undo what was just done. – robjohn Jul 04 '13 at 10:09
  • Thanks alot for your great effort! – TheNotMe Jul 04 '13 at 10:54
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I'd start using a geometric series, let $|x|<1$ then: $$\sum_{n=1}^\infty (-1)^n x^{2n}=-\frac{x^2}{1+x^2}=\frac{1}{1+x^2}-1.$$ Then by integrating with respect to $x$ and dividing by $x^2$ you get robjohn's result

Jason Hancox
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