I am trying to find a tight bound on the following sum: \begin{align} \sum_{i=1}^n \sum_{k=1}^i \frac{1}{(n-k+1)^2} \end{align}

If we use Wolfram_Alpha, we can actually find a closed-form expression: \begin{align} \sum_{i=1}^n \sum_{k=1}^i \frac{1}{(n-k+1)^2}=\gamma+\psi(1-n)+\frac{1}{n} \end{align} where $\gamma$ is Euler constant and $\psi$ is the digamma function. The exact result is a bit confusing to me. Specifically, I never seen the digamma function for negative integers.


1 Answer 1


Hint : $$\sum_{i=1}^n \sum_{k=1}^i \frac{1}{(n-k+1)^2} = \sum_{k=1}^n \sum_{i=k}^n \frac{1}{(n-k+1)^2} = \sum_{k=1}^n \frac{n-k+1}{(n-k+1)^2} = \sum_{k=1}^n \frac{1}{n-k+1}$$

  • $\begingroup$ Can you explain the first step? The rest I follow $\endgroup$
    – Lisa
    Commented Aug 28, 2020 at 16:26
  • $\begingroup$ I interverted the two summations, changing the indices : $\sum_{i=1}^n \sum_{k=1}^i a_{ik} = \sum_{k=1}^n \sum_{i=k}^n a_{ik}$. $\endgroup$ Commented Aug 28, 2020 at 16:33

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