For $\alpha>0$, let \begin{equation*} H_m^{(\alpha)}=\sum_{k=1}^m\frac{1}{k^{\alpha}} \end{equation*} What is $ \sum_{m=1}^n H_m^{(\alpha)} ? $ Does simply identity exist?
sum of generalized harmonic numbers
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number-theory
special-functions
harmonic-analysis
harmonic-functions
1 Answers
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It's $$\sum_{m=1}^{n} H_{m}^{(\alpha)}=\sum_{k=1}^{n}\frac{n-k+1}{k^\alpha} = (n+1)H_n^{(\alpha)}-H_{n}^{(\alpha-1)}$$
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0May I ask how you obtain the first equality? – 2017-01-15
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0How many times does $\frac{1}{k^\alpha}$ occur in the left hand sum? It occurs once per every $m$ with $k\leq m\leq n$. That means $n-k+1$ occurrences. – 2017-01-15
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0Thanks a lot! Appreciated. – 2017-01-15