I'm trying to take that sum: $$\sum_{k=1}^n H_{n+k}$$ So I transformed this sum to such: $\sum_{i=1}^n iH_{2n+1-i}$, unfortunately i can't make this sum out :( Hope You can help me, Thanks for attention!


You could replace $H_{n+k}$ with its definition and then change the order of summation.

A little bit simpler way is to note that your sum is $S_{2n}-S_n$, where $$S_n=\sum_{k=1}^{n}H_k=\sum_{k=1}^{n}\sum_{j=1}^{k}\frac{1}{j}=\sum_{j=1}^{n}\sum_{k=j}^{n}\frac{1}{j}\\=\sum_{j=1}^{n}\frac{n+1-j}{j}=(n+1)H_n-n.$$

  • $\begingroup$ (Yet another way is to use summation by parts - it's funny but I see incorrect formulae in the Wikipedia article.) $\endgroup$ – metamorphy Jun 23 at 20:33
  • $\begingroup$ Thank You:))))) $\endgroup$ – Katy Jun 25 at 11:32

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