My question may seem to be too silly as I am not quite in touch with these things since long, for which I apologise in advance. How should I proceed to calculate the following? $$\sum_{n=0}^\infty\left(\frac{1}{n+1}\right)^4$$

  • $\begingroup$ This is a special case of $\sum_{n=1}^\infty \frac1{n^{2k}}$ where Bernoulli numbers and powers of $\pi$ are involved. $\endgroup$ – Jens Schwaiger Jan 17 '19 at 9:03
  • $\begingroup$ If you meant the infinite series: it is not a trivial sum: $$\sum_{n=1}^\infty\frac1{n^4}=\zeta(4)=\frac{\pi^4}{90}$$ I use, for example, Fourier series to do this. $\endgroup$ – DonAntonio Jan 17 '19 at 9:03
  • $\begingroup$ Duplicate: math.stackexchange.com/questions/650966/… $\endgroup$ – Damien Jan 17 '19 at 9:38
  • $\begingroup$ I'd like to propose another duplicate target, and older post (#28329) that has more content and more links. $\endgroup$ – Lee David Chung Lin Jan 17 '19 at 12:51

Just for your curiosity.

As you see from the comments, the problem is not simple since involving for the infinite sum the zeta function.

Still worse, for the partial sums $$S_p=\sum_{n=0}^p \frac 1{(n+1)^4}=\frac{\pi ^4}{90}-1-\frac{1}{6}\psi ^{(3)}(p+2)$$ where appears the polygamma function.

Fortunately, when $p$ is large, we can use asymptotics and get $$S_p\simeq \left(\frac{\pi ^4}{90}-1\right)-\frac{1}{3 p^3}+\frac{3}{2 p^4}+O\left(\frac{1}{p^5}\right)$$

Use your calculator for $p=5$. The exact result would be $\frac{1051361}{12960000}\approx 0.0811235$ while the above expansion would give $ \frac{\pi ^4}{90}-\frac{3751}{3750}\approx 0.0820566$ ($1.15$% relative error).

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