I was just wondering wether one can derive a closed form for $$\sum_{n=1}^{\infty}\frac{1}{\Gamma\left(\frac{1}{n}\right)\zeta\left(1+\frac{1}{n}\right)}$$ Numerical simulation gives $S=1.20154...$

The product formula between $\Gamma$ and $\zeta$ gives $$\sum_{n=1}^{\infty}\frac{1}{n\cdot a_n} \text{ with } a_n=\int_{0}^{\infty}\frac{x^{\frac{1}{n}}}{e^x-1}\text{d}x$$ But I can't seem to go any further.

Any suggestion ? Is it even doable ?

  • $\begingroup$ $S=1.20652522381311998892341483649465039450107654831603\ldots$ $\endgroup$ – metamorphy Dec 29 '18 at 9:52

Your Answer

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

Browse other questions tagged or ask your own question.