This is from an Olympiad preparation course. We have to calculate the sum of the following series:

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

I do not really know where to start, I thought about using Euler's and De Moivre's identities to work with exponentials, but I didn't get far. Wolfram evaluates the series to be equal to $1$.

Are there some general strategies in evaluating series like this or do I just try to work my way around with some complex number identity?

Thanks in advance!

  • 3
    $\begingroup$ Are you sure that's not supposed to be $\pi(n-1)$ instead of $\pi n - 1$ ? $\endgroup$
    – K.defaoite
    Oct 15 '20 at 14:49
  • $\begingroup$ Yes, it is definately $\frac{\pi n-1}{2}$. However, I would be interested in the way you would solve it if it were $\pi(n-1)$, as I'm looking to understand the general ideas and not just the solution to this example. I seem to understand, that you may "sum $\pi$'s if that were the case, but I don't see how this can help here. Cheers! $\endgroup$
    – Summand
    Oct 15 '20 at 14:53
  • 1
    $\begingroup$ Using the cosine addition formula you can split this into two sums $$\mathcal{S}=\sin(1/2)\sum_{n=0}^\infty \frac{\sin(n\pi/2)}{2^n n!}+\cos(1/2)\sum_{n=0}^\infty \frac{\cos(n\pi/2)}{2^n n!}$$ And then use that $\cos(n\pi/2)$ goes $1,0,-1,0....$ and $\sin(n\pi/2)$ goes $0,1,0,-1....$. I think this will lead to nicer sums. $\endgroup$
    – K.defaoite
    Oct 15 '20 at 15:01

\begin{align} \sum_{n=0}^{\infty}{\frac{1}{n!2^{n}}\cos\left({\frac{\pi n-1}{2}}\right)}&=\frac12\sum_{n=0}^{\infty}{\frac{1}{n!2^{n}}\left(e^{i\frac{\pi n-1}2}+e^{-i\frac{\pi n-1}2}\right)}\\ &=\frac12\sum_{n=0}^{\infty}{\frac{1}{n!2^{n}}e^{i\frac{\pi n-1}2}}+\frac12\sum_{n=0}^{\infty}{\frac{1}{n!2^{n}}e^{-i\frac{\pi n-1}2}}\\ &=\frac1{2e^{i/2}}\sum_{n=0}^{\infty}{\frac{1}{n!2^{n}}e^{i\frac{\pi n}2}}+\frac{e^{i/2}}2\sum_{n=0}^{\infty}{\frac{1}{n!2^{n}}e^{-i\frac{\pi n}2}}\\ &=\frac1{2e^{i/2}}e^{e^{\pi i/2}/2}+\frac{e^{i/2}}2e^{e^{-\pi i/2}/2}\\ &=1.\\ \end{align}

Alternatively, since $\cos(\frac{\pi n-1}2)=\cos(\frac12)\cos(\frac{\pi n}2)+\sin(\frac12)\sin(\frac{\pi n}2)$,

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


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