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Find $$\sum_{k=1}^n\frac{\left(H_k^{(p)}\right)^2}{k^p}\,,$$ where $H_k^{(p)}=1+\frac1{2^p}+\cdots+\frac1{k^p}$ is the $k$th generalized harmonic number of order $p$.

Cornel proved in his book, (almost) impossible integral, sums and series, the following identity :

$$\sum_{k=1}^n\frac{\left(H_k^{(p)}\right)^2}{k^p}=\frac13\left(\left(H_n^{(p)}\right)^3-H_n^{(3p)}\right)+\sum_{k=1}^n\frac{H_k^{(p)}}{k^{2p}}$$

using series manipulations and he also suggested that this identity can be proved using Abel's summation and I was successful in proving it that way. other approaches are appreciated.

I am posting this problem as its' importance appears when $n$ approaches $\infty$.

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using Abel's summation $\ \displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)$ where $\displaystyle A_n=\sum_{i=1}^n a_i$

letting $\ \displaystyle a_k=\frac{1}{k^p}$ and $\ \displaystyle b_k=\left(H_k^{(p)}\right)^2$, we get \begin{align} S&=\sum_{k=1}^n\frac{\left(H_k^{(p)}\right)^2}{k^p}=\sum_{i=1}^n\frac{\left(H_{n+1}^{(p)}\right)^2}{i^p}+\sum_{k=1}^n\left(\sum_{i=1}^k\frac1{i^p}\right)\left(\left(H_k^{(p)}\right)^2-\left(H_{k+1}^{(p)}\right)^2\right)\\ &=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}+\sum_{k=1}^n\left(H_k^{(p)}\right)\left(\left(H_k^{(p)}\right)^2-\left(H_{k+1}^{(p)}\right)^2\right)\\ &=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}+\sum_{k=1}^{n+1}\left(H_{k-1}^{(p)}\right)\left(\left(H_{k-1}^{(p)}\right)^2-\left(H_{k}^{(p)}\right)^2\right)\\ &=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}-\sum_{k=1}^{n+1}\left(H_{k-1}^{(p)}\right)\left(\frac{2H_k^{(p)}}{k^p}-\frac1{k^{2p}}\right)\\ &=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}-\sum_{k=1}^{n}\left(H_{k}^{(p)}-\frac1{k^p}\right)\left(\frac{2H_k^{(p)}}{k^p}-\frac1{k^{2p}}\right)-\left(H_{n}^{(p)}\right)\left(\frac{2H_{n+1}^{(p)}}{(n+1)^p}-\frac1{{(n+1)}^{2p}}\right)\\ &=\underbrace{\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}-\left(H_{n}^{(p)}\right)\left(\frac{2H_{n+1}^{(p)}}{(n+1)^p}-\frac1{{(n+1)}^{2p}}\right)}_{\Large\left(H_n^{(p)}\right)^3}-2S+3\sum_{k=1}^n\frac{H_k^{(p)}}{k^{(2p)}}-H_n^{(3p)}\\ &=-2S+3\sum_{k=1}^n\frac{H_k^{(p)}}{k^{(2p)}}+\left(H_n^{(p)}\right)^3-H_n^{(3p)} \end{align}

which follows $$S=\frac13\left(\left(H_n^{(p)}\right)^3-H_n^{(3p)}\right)+\sum_{k=1}^n\frac{H_k^{(p)}}{k^{2p}}$$

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  • $\begingroup$ Keeping track of the summation indices it seems like there is small typo or error while going from the upper bound $n$ to $n+1$ as the lower bound remains uneffected afterall. Is this intended, since I cannot make sense out it right now. $\endgroup$ – mrtaurho Jun 26 at 7:29
  • $\begingroup$ @mrtaurho yes its intended as the index can start from zero instead of one then i added one to the lower and upper limit. $\endgroup$ – Ali Shather Jun 26 at 7:32
  • $\begingroup$ I see. However, very interesting solution! I really like the concept of SBP (i.e. Abel' Summation) but know way to few examples where its usage is actually helpful; this is one certainly is such an example! (+1) $\endgroup$ – mrtaurho Jun 26 at 7:36
  • $\begingroup$ thank you and i agree its not always helpful. by the way you can use the same approach to prove $$\sum_{k=1}^n\frac{H_k^{(p)}}{k^p}=\frac12\left((H_n^{(p)})^2+H_n^{(2p)}\right)$$ $\endgroup$ – Ali Shather Jun 26 at 8:44

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