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Find sum of the series $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+3)(2n+1)}.$$

I think I have to find known taylor series and modify it to look like the above but I can't see which one to pick.

The solution is $\frac{\pi -2}{4}$ so I'm guessing it has to be one of the trig functions.

I picked $\sin x$ $$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

integrate $$-\cos(x)=\sum_{n=0}^\infty \frac{(-1)^{n} x^{2n+2}}{(2n+2)(2n+1)!}.$$

The index is not correct plus I don't think I can get it to look like the problem.

Can someone explain which function to pick and how to solve?

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  • $\begingroup$ Might this help: $\dfrac1{(2n+3)(2n+1)} = \dfrac12\left[\dfrac1{2n+1}-\dfrac1{2n+3}\right]$ $\endgroup$ – DHMO Apr 17 '17 at 3:51
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$$\begin{array}{rcl} \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} &=& \displaystyle \sum_{n=0}^\infty \int_0^1 (-1)^n x^{2n} \ \mathrm dx \\ &=& \displaystyle \sum_{n=0}^\infty \int_0^1 (-x^2)^n \ \mathrm dx \\ &=& \displaystyle \int_0^1 \sum_{n=0}^\infty (-x^2)^n \ \mathrm dx \\ &=& \displaystyle \int_0^1 \dfrac1{1+x^2} \ \mathrm dx \\ &=& \displaystyle \left(\arctan x\right)_0^1 \\ &=& \displaystyle \dfrac\pi4 \\ \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)(2n+3)} &=& \dfrac12 \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} - \dfrac12\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+3} \\ &=& \dfrac12 \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} + \dfrac12\displaystyle \sum_{n=1}^\infty \dfrac{(-1)^n}{2n+1} \\ &=& \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} - \dfrac12 \\ &=& \dfrac\pi4 - \dfrac12 \\ \end{array}$$

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Maybe you will like this method. Let $$ f(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+3)(2n+1)}x^{2n+3}.$$ Then $$ f'(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+2}, (\frac{f(x)}{x})'=\sum_{n=0}^\infty(-1)^nx^{2n}=\frac{1}{1+x^2}.$$ So $$ f(1)=\int_0^1x\int_0^x\frac{1}{1+t^2}dtdx=\int_0^1x\arctan xdx=\frac{\pi}{4}-\frac12. $$

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