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I am interested in the rational zeta series as title. WolframAlpha gives the result: $$ \sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1}=0.78853056591150896106... $$ This is the Lüroth analog of Khintchine’s constant, which is defined as: $$ \sum_{n=1}^{\infty}\frac{\ln (n)}{n(n+1)}=-\sum_{n=2}^{\infty}(-1)^{n}\zeta^{'}(n) $$ Why this rational zeta series is not in terms of $\gamma$, $\ln2$, $\ln \pi$ or normal constant that usually seen, but comes out an unusual constant? How to evaluate it to a closed form?

Also how to evaluate its companion zeta series to a closed form? $$ \sum_{n=2}^{\infty}\frac{(-1)^{n}\left ( \zeta(n)-1 \right )}{n-1}=0.56459970638442432059... $$ Or there are no closed forms for both zeta series?

Nanhui Lee
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  • Why would you sysoect a closed form exists? – Mark Viola Jul 12 '20 at 01:41
  • I am not sure if a closed form exists. It's a relatively simple form zeta series, and should have been studied in depth. That's why I asked this question to see if some experts know its closed form, or know that there is no closed form. – Nanhui Lee Jul 12 '20 at 01:51
  • Series much simpler than this have no (known) closed form, Nanhui, e.g., $\sum^{\infty}n^{-3}$. – Gerry Myerson Jul 12 '20 at 01:58
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    The first series is discussed at http://oeis.org/A085361 where there is a link to 1,000 decimal places & some other information but no closed form is given (which strongly suggests none is known). – Gerry Myerson Jul 12 '20 at 02:01
  • Thank you very much for your information. – Nanhui Lee Jul 12 '20 at 02:18
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    There is a simple rule (works most of the time) to determine whether such series have a "closed form": for infinite sum $$\sum_n \frac{\log^k(n+b)}{(n+a)^s}$$ there is no simple answer when $a\neq b$. In your case, you have $\sum_n \log n / (1+n)$, this precludes nice closed-form. You need to resort to some peculiar function like poly-Hurwitz zeta for this. – pisco Aug 04 '20 at 05:56

1 Answers1

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Although there are no closed forms for both rational zeta series, we can evaluate them with definite integrals.

Set $f(x)$ as \begin{align*} f(x)&=\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1}x^{n-1}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{k^n}\int_{0}^{x}t^{n-2}\,dt=\int_{0}^{x}\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{t^{n-2}}{k^n}\,dt \\ &=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\sum_{n=2}^{\infty}\left( \frac{t}{k} \right)^{n-2}dt=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\frac{1}{\left(1-\frac{t}{k}\right)}\,dt \\ &=\int_{0}^{x}\left(\frac{1}{t-1}+\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{(k-t)} \right)dt=\int_{0}^{x}\frac{1}{t-1}\,dt-\int_{0}^{x}\frac{\gamma+\psi(1-t)}{t}\,dt \\ &=\ln(1-x)+\left [ \frac{\ln\Gamma(1-t))}{t} \right ]_{t=0}^{t=x}+\int_{0}^{x}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\ &=\ln(1-x)+\frac{\ln\Gamma(1-x)}{x}-\gamma+\int_{0}^{x}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\ \\ f(1)&=\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1} \\ &=\lim_{x\rightarrow 1^{-}}\left ( \ln(1-x))+\frac{\ln\Gamma(1-x))}{x} \right )-\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\ &=-\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \approx0.7885306 \\ \end{align*} In the same way, set $f(x)$ as \begin{align*} f(x)&=\sum_{n=2}^{\infty}\frac{(-1)^{n}(\zeta(n)-1)}{n-1}x^{n-1}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{(-1)^{n}}{k^n}\int_{0}^{x}t^{n-2}\,dt=\int_{0}^{x}\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{(-t)^{n-2}}{k^n}\,dt \\ &=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\sum_{n=2}^{\infty}\left( -\frac{t}{k} \right)^{n-2}dt=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\frac{1}{\left(1+\frac{t}{k}\right)}\,dt \\ &=\int_{0}^{x}\left(-\frac{1}{t+1}+\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{(k+t)} \right)dt=-\int_{0}^{x}\frac{1}{t+1}\,dt+\int_{0}^{x}\frac{\gamma+\psi(1+t)}{t}\,dt \\ &=-\ln(1+x)+\left [ \frac{\ln\Gamma(1+t))}{t} \right ]_{t=0}^{t=x}+\int_{0}^{x}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \\ &=-\ln(1+x)+\frac{\ln\Gamma(1+x)}{x}+\gamma+\int_{0}^{x}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \\ \\ f(1)&=\sum_{n=2}^{\infty}\frac{(-1)^{n}(\zeta(n)-1)}{n-1} \\ &= -\ln(2)+\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \approx0.5645997 \\ \end{align*}

Nanhui Lee
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