1

I encountered the following sum in my work and I was wondering if it has a known closed form: $$ \sum^{\infty}_{n=1} \frac{1}{(n+1)n^{\alpha}} \quad , \quad 0 < \alpha < 1 \; , \; \alpha \in \mathbb{Q} $$ It is convergent for all $\alpha$ by the p-series test, and I have been able to find an alternative formulation in terms of Riemann Zeta function by using the asymtotic expansion of $1 / (n+1)$. $$ \sum^{\infty}_{n=1} \frac{1}{(n+1)n^{\alpha}} = \sum^{\infty}_{n,m=1} \frac{(-1)^{m+1}}{n^{m+\alpha}} = \sum^{\infty}_{m=1} (-1)^{m+1}\zeta (m+\alpha) $$ However I'm dubious of this transformation because it fails the alternating series test, but I'm unsure on what exactly what the error in this representation is. This differs from similar problems where $\alpha$ is taken to be an integer as it being rational between 0 and 1 the same telescoping of the Zeta functions can't occur.

Lorenzo B.
  • 2,290
Aidan R.S.
  • 1,042
  • 1
    @Zima for integer order polylogorithms you are able to get a closed for in the form of a finite sum of zeta functions. For non-integer values however the same cancelations cant occur. I tried to see if there was a way to make those same cancelations happen but I wasn't able to find a way to seperate the integer and non-integer parts of the evaluation of the zeta function losely speaking. – Aidan R.S. Jul 07 '24 at 10:21
  • 1
    Can be expressed by integral: $$\sum _{n=1}^{\infty } \frac{1}{(n+1) n^{\alpha }}=\int_0^{\infty } \frac{(-1)^{-\alpha } e^{-x} (\Gamma (\alpha )-\Gamma (\alpha ,-x))}{\Gamma (\alpha ) (\exp (x)-1)} , dx$$ – Mariusz Iwaniuk Jul 07 '24 at 12:05
  • @MariuszIwaniuk, how is that derived? – Simon Jul 07 '24 at 12:27
  • 3
    For $\alpha = 1/2$ this is the Theodorus constant – jjagmath Jul 07 '24 at 13:12
  • 2
    Also, $$\sum _{n=1}^\infty\frac1{(n+1) n^\alpha }=\frac1{\Gamma(1+\alpha)}\int_0^\infty\left(\ln(1-e^{-t})+\frac1{e^t-1}\right)t^\alpha e^tdt$$ $$=\frac1{\Gamma(1+\alpha)}\int_0^\infty\Big(x-\ln(1+x)\Big)\ln^\alpha\left( \frac {1+x}x\right)\frac{dx}{x^2}$$ – Svyatoslav Jul 07 '24 at 16:07

1 Answers1

5

I don't think there is a closed form for such a series, it looks too complicated. However, your formula can't be true because $\zeta(s) \rightarrow 1$ when $s \rightarrow +\infty$ hence the series diverges. For all $n \geqslant 2$, we have $-1 < \frac{1}{n} < 1$, hence, $$ \frac{1}{(n + 1)n^\alpha} = \frac{1}{n^{\alpha + 1}}\frac{1}{1 + 1/n} = \frac{1}{n^{\alpha + 1}}\sum_{m \geqslant 0} \frac{(-1)^m}{n^m}. $$ However, this is false for $n = 1$ because the right hand side diverges in this case (this is probably what you missed). When $n \geqslant 2$, everything converges absolutely so we can exchange the sum symbols to obtain that, \begin{align*} \sum_{n \geqslant 1} \frac{1}{(n + 1)n^\alpha} & = \frac{1}{2} + \sum_{n \geqslant 2} \frac{1}{(n + 1)n^\alpha}\\ & = \frac{1}{2} + \sum_{m \geqslant 0}\sum_{n \geqslant 2} \frac{(-1)^m}{n^{m + \alpha + 1}}\\ & = \frac{1}{2} + \sum_{m \geqslant 0}(-1)^m\left(-1 + \sum_{n \geqslant 1} \frac{1}{n^{m + \alpha + 1}}\right)\\ & = \frac{1}{2} + \sum_{m \geqslant 1}(-1)^{m + 1}(\zeta(m + \alpha) - 1). \end{align*} This time, the right hand side converges absolutely because $\zeta(s) - 1 \sim \frac{1}{2^s}$ when $s \rightarrow +\infty$. All of this still holds for any real $\alpha > 0$ with the same arguments (and even any complex $\alpha$ with $\Re(\alpha) > 0$).

Cactus
  • 9,220