# Evaluate infinite sum for $\frac{1}{n^4}$ using integration

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4}$

I want to evaluate this sum by the use of integration.

$\displaystyle \int_{\frac{1}{n^4}}^{\frac{4}{n^4}} 1 \space dx = \frac{4}{n^4} - \frac{1}{n^4} = \frac{3}{n^4}$

$\displaystyle \int_{1}^{\infty} 1 \space dx = \sum_{n=1}^{\infty} \int_{\frac{1}{n^4}}^{\frac{4}{n^4}} 1 \space dx$

The improper integral does not converge. So what is a way of finding this sum using integrals??

• this does not make any sense. – mookid Oct 31 '14 at 21:16
• @mookid, what part? The question or the workings? If the workings then perhaps you could help. – Amad27 Oct 31 '14 at 21:16
• the workings. You can't evaluate to a close form the integral using integration. – mookid Oct 31 '14 at 21:17
• You could always write $\sum_{n=1}^\infty \frac{1}{n^4} = \int_1^\infty \frac{dx}{\lfloor x \rfloor^4}$, but that would not make it any easier to solve. In general integrals are used to approximate sums, and its only in some cases like $\int_0^1 x^{-x} dx = \sum_{n=1}^\infty n^{-n}$ that one has a direct relation. For this problem I don't think there is such a (simple) relation that allows you to evaluate it. But lets hope someone proves me wrong:) – Winther Oct 31 '14 at 21:18
• Why do you think the intervals $[1/n^4,4/n^4]$ are a simply covering of $[1,\infty]$? Seems like $[1/16,1/4]$ and $[1/81,4/81]$ overlap. Also, the maximum value is $4/1^4=4$. Really not clear how you get that sum in the last eqality. – Thomas Andrews Oct 31 '14 at 21:18

Your last equation isn't true, which is probability related to why it's not working. In particular, notice that, if we start writing out the terms of your sum on the left hand side, we get: $$\left(\int_{1}^{4}1\,dx\right)+\left(\int_{1/16}^{1/4}1\,dx\right)+\left(\int_{1/81}^{4/81}1\,dx\right)+\cdots$$ This poses a problem for your equality, since you're integrating, essentially, over the union of all intervals of the form $[\frac{1}{n^4},\frac{4}{n^4}]$, and your equality assumes that this union is $[1,\infty)$, which is a problem, because, from the above,it is clear that the union of such intervals is, in fact, not connected and is bounded. So, your equality doesn't work.
It is well known that $$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$$ This value is not obtained by integration in the manner in which you are trying. In fact, I have no idea what you are attempting to do but it seems wrong. This question is much more difficult than just an integral. I suggest reading about "Particular values of Riemann zeta function" to get an idea of how you would evaluate a sum like this. http://en.wikipedia.org/wiki/Particular_values_of_Riemann_zeta_function
$$\sum_{n=1}^{\infty}\frac{1}{n^4} < 1 + \int_{1}^{\infty} \frac{1}{x^4} dx$$
$$\sum_{n=1}^{\infty}\frac{1}{n^4} = \int_{1}^{\infty} \frac{1}{\lfloor x\rfloor^4} dx$$