I am trying to show that the following statement holds for every $k\in\mathbb{N}^*$:
$$\sum_{n=0}^{k-1}\frac{2n+2-k}{(2n+2-k)^2+k^2}=\frac1{2k}$$
The original proof is via an extremely long work with digamma functions and complex numbers. For this reason I wondered if a more straightforward proof can be obtained withouth imaginary numbers and special functions.
For this reason I thought induction would do the job. However, when dealing with the inductive step, I supposed that for some $k\ge 1$ the equality holds, and I looked at the $k+1$ case.
Here comes the issue: plugging in $k+1$ instead of $k$ turns the summand in something that is not related to the previous case (apart from having $k+1$ instead of $k$)
\begin{align}
& k: &&& \frac{2n+2-k}{(2n+2-k)^2+k^2} \\
\\
& k+1: &&& \frac{2n+1-k}{(2n+1-k)^2+(k+1)^2} \\
\end{align}
so I couldn't use the inductive hypothesis, since it is a different sum from before.
To make the matter clear, we know that
$$\sum_{n=0}^{k-1}\frac{2n+2-k}{(2n+2-k)^2+k^2}=\frac1{2k}$$
but I can't see how to use this fact in the evaluation of
$$\sum_{n=0}^{k}\frac{2n+1-k}{(2n+1-k)^2+(k+1)^2}$$
since the denominators are different.
Any ideas on how to prove the statement?
Best Answer
Firstly let remove $+2$ by increasing $n$ by $1$: $$\sum_{n = 0}^{k - 1}\frac{2n + 2 - k}{(2n + 2 - k)^2 + k^2} = \sum_{n = 1}^{k}\frac{2n - k}{(2n - k)^2 + k^2}.$$ Now note that $$\frac{2n - k}{(2n - k)^2 + k^2} + \frac{k - 2n}{(k - 2n)^2 + k^2} = 0.$$ Therefore we can match each summand with negative numerator to its positive sibling, which annihilates it. Also there is one positive summand which doesn't match to any negative one, and for even $k$ there is one summand with zero numerator: $$\begin{aligned}&\sum_{n = 1}^{k}\frac{2n - k}{(2n - k)^2 + k^2}\\ =& \sum_{n = 1}^{\left\lfloor\frac{k - 1}{2}\right\rfloor}\frac{2n - k}{(2n - k)^2 + k^2} + [2 \mid k]\cdot \frac{2\frac{k}{2} - k}{(2\frac{k}{2} - k)^2 + k^2} + \sum_{n = \left\lceil\frac{k + 1}{2}\right\rceil}^{k - 1}\frac{2n - k}{(2n - k)^2 + k^2} + \frac{2k - k}{(2k - k)^2 + k^2}\\ =& \sum_{n = 1}^{\left\lfloor\frac{k - 1}{2}\right\rfloor}\frac{2n - k}{(2n - k)^2 + k^2} + \sum_{n = 1}^{\left\lfloor\frac{k - 1}{2}\right\rfloor}\frac{2(k - n) - k}{(2(k - n) - k)^2 + k^2} + \frac{k}{2k^2}\\ =& \sum_{n = 1}^{\left\lfloor\frac{k - 1}{2}\right\rfloor}\left(\frac{2n - k}{(2n - k)^2 + k^2} + \frac{k - 2n}{(k - 2n)^2 + k^2}\right) + \frac{1}{2k}\\ =& \frac{1}{2k}. \end{aligned}$$