Computing a partial sum and finding its sequence Let $S=\sum_{n=1}^{\infty}a_n$ be an infinite series such that its partial sum is given by $S_n=8-\frac{2}{n^2}$.

*

*Find $\sum_{n=1}^{10}a_n$ and $\sum_{n=4}^{16}a_n$

*Find a general formula for $a_n$

I found the first part of (1). I computed $S_{10}$ to get $8-1/50$. I'm struggling with the second part of (1). I figured it was $S_{16}-S_4$, but that didn't work.
For problem (2) I found a couple terms of the sequence, $$6, \frac32, \frac{5}{18}, \frac{7}{72}, \cdots$$I know that $S=\lim_{n\rightarrow \infty}S_n=8$. Assuming this is correct, I see no pattern here to get the sequence.
 A: Remember this well-known formula:

*

*$S_n-S_{n-1}=a_n$
Then use: $S_n=\sum_{i=1}^{n}a_i$

If you ask, where does this formula come from?
It comes from here:
$$ \begin{align}S_n=\underbrace {a_1+a_2+a_3+\cdots +a_{n-1}}_{S_{n-1}} +a_n =S_{n-1}+a_n \end{align}$$
$$ \implies a_n= S_n-S_{n-1}$$
A: By definition of partial sums, $$S_n=\sum_{i=1}^n a_n$$ therefore by definition $$\sum_{n=1}^{10}a_n=S_{10}=8-\frac{1}{50} \text{ and } \sum_{n=4}^{16}a_n=S_{16}-S_3.$$
To find $a_n$, note that $a_n=S_{n}-S_{n-1}$.
A: To be step-by-step explicit for the second part of (1), we have
$$\begin{align}
\sum_{n=4}^{16}a_n
&=a_4+a_5+\cdots+a_{16}\\
&=(a_1+a_2+a_3+a_4+a_5+\cdots+a_{16})-(a_1+a_2+a_3)\\
&=\sum_{n=1}^{16}a_n-\sum_{n=1}^3a_n\\
&=S_{16}-S_3\\
&=\left(8-{2\over16^2}\right)-\left(8-{2\over3^3}\right)\\
&={2\over9}-{1\over128}\\
&={247\over1152}
\end{align}$$
As for part (2), the OP is quite right that there is no obvious pattern that accounts for the first four values of $a_n$. This is because there isn't one! As others have pointed out, there is a general formula,
$$a_n=S_n-S_{n-1}=\left(8-{2\over n^2}\right)-\left(8-{2\over(n-1)^2}\right)=2\left({1\over(n-1)^2}-{1\over n^2}\right)={2(2n-1)\over n^2(n-1)^2}$$
But this formula does not apply to $n=1$. For $n=1$, we simply have $a_1=S_1=8-{2\over1^2}=6$, so the answer comes in two pieces:
$$a_n=\begin{cases}
6&\text{if }n=1\\
\displaystyle{2(2n-1)\over n^2(n-1)^2}&\text{if }n\gt1
\end{cases}$$
