$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4}$
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??
On
What you have is
$$\sum_{n=1}^{\infty}\frac{1}{n^4} < 1 + \int_{1}^{\infty} \frac{1}{x^4} dx$$
or
$$\sum_{n=1}^{\infty}\frac{1}{n^4} = \int_{1}^{\infty} \frac{1}{\lfloor x\rfloor^4} dx$$
On
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
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.