# Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series

The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of $$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$

Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:

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

So I wrote down the partial sums of the given series and $\sin(1)$ and $\cos(1)$:

$$\qquad\qquad\quad\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \quad - \frac{1}{4!} + \frac{2}{6!} - \frac{3}{8!} \cdots$$

$$\;\;\sin(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n+1}}{(2n+1)!} = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots$$

$$\cos(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n}}{(2n \qquad)!} = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} \cdots$$

Looking at the numbers I can see that Wolfram|Alpha's result is correct: $\frac{1}{2}1 - \frac{1}{2!} = 0$ and $\frac{1}{2}\frac{-1}{3!} + \frac{1}{4!} = \frac{1}{4!}$, so the $\cos(1)$-series is shifted by $1$ since there is no $1$ at the beginning of the given series, so it needs to be subtracted from $\cos(1)$: $-cos(0)=-1$. But how do I get here without Wolfram|Alpha?

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No idea how Wolfram does it. But I guess you just start with power series and play around. $$X=\sum_{n=1}^\infty(-1)^n\frac{n}{(2n+2)!}=-\frac{1}{4!}+\frac{2}{6!}-\frac{3}{8!}+\dots$$ The annoying part is the numerators. Try doubling it so that the numerators "keep up" with the denominators. $$2X=-\frac{2}{4!}+\frac{4}{6!}-\frac{6}{8!}+\dots$$ Now the numerators are consistently off the denominators by 2. \begin{equation*} 2X=(-\frac{4}{4!}+\frac{6}{6!}-\frac{8}{8!}+\dots)+2(\frac{1}{4!}+\frac{1}{6!}-\frac{1}{8!}+\dots) \end{equation*} That second part is pretty much $\cos(1)$. And the numerators of course cancel out in the first part. $$2X=(-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\dots)+2(\cos(1)-1/2)$$ Now the first part is pretty much $\sin(1)$. $$2X=(\sin(1)-1)+2\cos(1)-1$$ $$X=\frac{\sin(1)}{2}+\cos(1)-1$$

Hope that didn't seem too random. My approach is to try to build the desired series out of ones I know using various manipulations like calculus, combining and rearranging.

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Thanks. I chose this one as answer because it carried on my approach. –  mcb Jan 5 '12 at 7:49

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

and

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

The given series has factorials of even numbers in the denominator, so I began by shifting indices to make them $(2n)!$ to match the cosine series. After that it was mostly a matter of following my nose: at each step in the calculation below there’s really only one thing that suggests itself strongly.

\begin{align*} \sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}&=\sum_{n=2}^\infty(-1)^{n-1}\frac{n-1}{(2n)!}\\ &=\sum_{n=2}^\infty(-1)^{n-1}\frac{n}{(2n)!}-\sum_{n=2}^\infty(-1)^{n-1}\frac1{(2n)!}\\ &=\sum_{n=2}^\infty(-1)^{n-1}\frac{n}{2n(2n-1)!}+\sum_{n=2}^\infty\frac{(-1)^n}{(2n)!}\\ &=\frac12\sum_{n=2}^\infty\frac{(-1)^{n-1}}{(2n-1)!}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}-1+\frac12\\ &=\frac12\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)!}+\cos 1-\frac12\\ &=\frac12\left(\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}-1\right)+\cos 1-\frac12\\ &=\frac12\sin 1 + \cos 1 - 1\\ &=\frac12\sin 1+\cos 1-\cos 0 \end{align*}

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Thank you, that helped a lot! –  mcb Jan 5 '12 at 6:31

Of course you know the series for $\cos(1)$ and $\sin(1)$, so you want to express this using those.
I prefer to start the sum at $n=0$: since the $n=0$ term is $0$, this is harmless. Note that $\frac{n}{(2n+2)!} = \frac{n}{(2n+1)!(2n+2)}$. Now $\frac{n}{2n+2} = \frac{1}{2} - \frac{1}{2n+2}$. So $$\sum_{n=0}^\infty (-1)^n \frac{n}{(2n+2)!} = \frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} - \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+2)!}$$ Now note that $\sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} = \sin(1)$ and $$\sum_{n=0}^\infty (-1)^n \frac{1}{(2n+2)!} = \frac{1}{2!} - \frac{1}{4!} + \frac{1}{6!} - \ldots = 1 - \cos(1)$$

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let $m=n+1$, so your series reduces to $$\sum_{m=2}^\infty(-1)^{m-1}\frac{m-1}{(2m)!}=-\sum_{m=2}^\infty(-1)^{m}\frac{m}{(2m)!}+\sum_{m=2}^\infty(-1)^{m} \frac{1}{(2m)!}$$ $$=-\sum_{m=2}^\infty(-1)^{m}\frac{1}{2(2m-1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$

Note that $\frac{m}{(2m)!}=\frac{m}{(2m)(2m-1)!}=\frac{1}{(2)(2m-1)!}$, and also we evaluated the second series at $m=0$ and $m=1$

Replacing $m$ by $k+1$ in the first series and simplifying, we get, $$\sum_{k=1}^\infty(-1)^{k}\frac{1}{2(2k+1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$ $$\sum_{k=0}^\infty(-1)^{k}\frac{1}{2(2k+1)!}-\frac{1}{2!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$ $$\sum_{k=0}^\infty(-1)^{k}\frac{1}{2(2k+1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1$$ $$=\frac{\sin (1)}{2}+\cos(1)-\cos(0)$$

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