# Sum of the following infinite series [duplicate]

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? $$\sum_{n=0}^\infty\left(\frac{1}{n+1}\right)^4$$

## marked as duplicate by Arnaud D., Lee David Chung Lin, RRL, A. Pongrácz, rtybaseJan 17 at 22:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• This is a special case of $\sum_{n=1}^\infty \frac1{n^{2k}}$ where Bernoulli numbers and powers of $\pi$ are involved. – Jens Schwaiger Jan 17 at 9:03
• 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. – DonAntonio Jan 17 at 9:03
• – Damien Jan 17 at 9:38
• I'd like to propose another duplicate target, and older post (#28329) that has more content and more links. – Lee David Chung Lin Jan 17 at 12:51

## 1 Answer

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).