How to calculate the following sum?

Question

$\sum _{n=1}^{\infty \:}\frac{10000}{\left(2n+3\right)^4}$

I could only prove that it is convergent, but I have no idea how to find the sum. Thanks for the help :-)

Answer 1

Call $S$ the sum.

We know that $$\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}$$

Then $$\sum_{n=1}^\infty\frac1{(2n)^4}=\frac{\pi^4}{1440}$$

Therefore $$\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac{\pi^4}{90}-\frac{\pi^4}{1440}=\frac{\pi^4}{96}$$

Thus, $$S=10000\left(\frac{\pi^4}{96}-1-\frac1{81}\right)$$

Answer 2

First, you can ignore the factor 10000, it is only here for show. Then, write $$\begin{align} \sum_{n=1}^\infty \frac{1}{n^4} &= \sum_{n=1}^\infty \frac{1}{(2n)^4} + \sum_{n=1}^\infty \frac{1}{(2n-1)^4} = \frac{1}{16}\sum_{n=1}^\infty \frac{1}{n^4} + \sum_{n=1}^\infty \frac{1}{(2n-1)^4} \\ &= \frac{1}{16}\sum_{n=1}^\infty \frac{1}{n^4} + 1 + \sum_{n=0}^\infty \frac{1}{(2n+3)^4} = \frac{1}{16}\sum_{n=1}^\infty \frac{1}{n^4} + 1 + \frac{1}{3^4} + \sum_{n=1}^\infty \frac{1}{(2n+3)^4} \end{align}$$ so that $\sum_{n=1}^\infty \frac{1}{(2n+3)^4} = -1 -\frac{1}{81}+ \frac{15}{16} \sum_{n=1}^\infty \frac{1}{n^4}$. Can you compute $\sum_{n=1}^\infty \frac{1}{n^4}$?