Find the sum (in closed form) of this series:

$$\sum_{k=1}^{\infty} \frac{\log\left(\cos\left(\sin \left(k\frac{\pi}{2}\right)\right)\right)}{k^2} $$

I have been given a hit to use properties of logarithms, but I don't know how to do that.

  • $\begingroup$ First compute $\{\sin(1\cdot \pi/2),\sin(2\cdot \pi/2), \ldots\} = \{1,0,-1,0,1,0,-1,0,\ldots\}$. Next compute $\cos$ of this sequence (using that $\cos(x) = \cos(-x)$). Then take the log (remember that $\log(1) = 0$) and finally you can use that $\sum \frac{1}{k^2} = \sum \frac{1}{(2k)^2} + \sum \frac{1}{(2k+1)^2}$ plus $\sum \frac{1}{k^2} = \frac{\pi^2}{6}$ to evaluate the series you are left with. $\endgroup$
    – Winther
    Commented Dec 29, 2015 at 21:19

1 Answer 1


Notice that the only possible value for $\sin$ are $0,1,-1$. Moreover, $\cos (0)=1$, and $\cos(-1) = \cos(1)$ since $\cos$ is even.

Hence the only values of $k$ that contribute to the sum are the odd ones.

So you have

$$\sum_{k=0}^{\infty} \frac{\log(\cos(1))}{(2k+1)^2}=\log(\cos(1))\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}$$

which is easy to evaluate (follow @Winther's hint).


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