Compute the series $\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}$ Compute the series $$\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}.$$
How do I go about with the index notation, for example to arrange the series instead as  $\sum_{n=1}^{\infty}a_n $?
I have tried to simplify the expression as:
$$\begin{align}&\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}=\sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies&\sum_{n=3}^{\infty} \frac{n^4}{n^6} - \sum_{n=3}^{\infty} \frac{1}{n^6} = \sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies& \sum_{n=3}^{\infty} \frac{1}{n^2} - \sum_{n=3}^{\infty} \frac{1}{n^6}\end{align}$$
I'm not sure what to do with the index $n=3$ as I know that I can simplify it otherwise as $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{\pi^6}{945}$.

$$
\begin{align}
\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}&=\sum_{n=3}^{\infty} \frac{n^4}{n^4}\frac{1-\frac{1}{n^4}}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1-\frac{1}{n^4}}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1}{n^2}-\frac{1}{n^6}\\
\end{align}$$
Now note, since each limit exists (see below, for the value)
$$\sum_{n=3}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}-\frac{5}{4}$$
$$\sum_{n=3}^{\infty} \frac{1}{n^6}=\frac{\pi^6}{945}-\frac{65}{64}$$
You can split up the sum
$$
\begin{align}
\sum_{n=3}^{\infty} \frac{1}{n^2}-\frac{1}{n^6}&=\sum_{n=3}^{\infty} \frac{1}{n^2}-\sum_{n=3}^{\infty}\frac{1}{n^6}\\
&=\frac{\pi^2}{6}-\frac{5}{4}-\frac{\pi^6}{945}+\frac{65}{64}\\
&=\frac{\pi^2}{6}-\frac{\pi^6}{945}-\frac{15}{64}\\
\end{align}$$
The calculation of the limits above is done in the following manner:
$$
\begin{align}
\frac{\pi^2}{6}&=\sum_{n=1}^{\infty} \frac{1}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1}{n^2}+\frac{1}{1^2}+\frac{1}{2^2}\\
&=\sum_{n=3}^{\infty} \frac{1}{n^2}+\frac{5}{4}\\
\end{align}$$
Subtracting $\frac{5}{4}$ on both sides, yields
$$
\begin{align}
\frac{\pi^2}{6}-\frac{5}{4}&=\sum_{n=3}^{\infty} \frac{1}{n^2}
\end{align}$$
