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?


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

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.

  • $\begingroup$ Thanks. I chose this one as answer because it carried on my approach. $\endgroup$ – Lenar Hoyt Jan 5 '12 at 7:49

Start with the known Maclaurin series

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


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

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.

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


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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.