How can I find partial sum of $$\sum\limits_{n=0}^{m}\frac{1}{na+b}$$ and infinite sum of $$\sum\limits_{n=0}^{\infty}\frac{1}{(na+b)^k}$$ if $k>1$ and $a,b$ - constants?
I sure it simple, but really have no ideas.
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I dont's think there there is a closed formula for this. – José Carlos Santos Jan 21 '18 at 12:32
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2A closed form for the latter, would be a closed form for the Riemann Zeta, when $a=1$ and $b=0$; thus I think it could be rather difficult – Joe Jan 21 '18 at 12:38
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As Joe already said the second sum is related to the Euler-Riemann Zeta function. Euler derived a closed form solution for $2k$ in which $k \in \mathbb{Z}- {0}$. – MrYouMath Jan 21 '18 at 12:52
2 Answers
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The closest form you will find ( I think ) is
$$ \sum_{n=0}^{m}\frac{1}{an+b}=\frac{1}{a}\left(\psi\left(m+\frac{a+b}{a}\right)-\psi\left(\frac{b}{a}\right)\right) $$
where $\psi$ is the digamma function given by $$ \psi\left(z\right)=\frac{\Gamma'\left(z\right)}{\Gamma\left(z\right)} $$
Atmos
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For the finite sums $$S_k=\sum\limits_{n=0}^{m}\frac{1}{(na+b)^k}=\frac {(-1)^k} {(k-1)!\, a^k}\left(\psi ^{(k-1)}\left(\frac{b}{a}\right)-\psi ^{(k-1)}\left(\frac{b}{a}+m+1\right) \right)$$ For the infinite sums (with $k>1)$, this simplifies to $$T_k=\sum\limits_{n=0}^{\infty}\frac{1}{(na+b)^k}=\frac {(-1)^k} {(k-1)!\, a^k}\,\psi ^{(k-1)}\left(\frac{b}{a}\right)$$ where appear the polygamma functions.
Claude Leibovici
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