1
$\begingroup$

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 $\infty$.

|cite|improve this question
$\endgroup$
2
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '19 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 '19 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 '19 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 '19 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.