Later data often show patterns better than early data, so extend your table of partial sums a bit:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171
\end{array}$$
Ignoring the signs, it appears that the numbers in the bottom line are approximately doubling each time. Moreover, still ignoring signs, adjacent partial sums add up to a power of $2$: $|s_1|+|s_2|=2^1$, $|s_2|+|s_3|=2^2$, $|s_3|+|s_4|=2^3$, and apparently in general $|s_n|+|s_{n+1}|=2^n$. (If you go back to the definition of the partial sums, you’ll see why this happens.)
If $|s_n|+|s_{n+1}|=2^n$ and $|s_{n+1}|\approx 2|s_n|$, then $3|s_n|\approx 2^n$; this suggests that we should compare $3|s_n|$ with $2^n$:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171\\
3|s_n|:&3&3&9&15&33&63&129&255&513\\
2^n:&2&4&8&16&32&64&128&256&512
\end{array}$$
That pattern’s pretty clear: apparently $3|s_n|=2^n+1$ if $n$ is odd, and $3|s_n|=2^n-1$ if $n$ is even. Those cases can be combined as $3|s_n|=2^n+(-1)^{n+1}$, since $(-1)^{n+1}$ is $1$ when $n$ is odd and $-1$ when $n$ is even. And the algebraic sign of $s_n$ appears to be that of $(-1)^{n+1}$, so if these patterns are real,
$$\begin{align*}
s_n&=\frac{(-1)^{n+1}}3\left(2^n+(-1)^{n+1}\right)\\
&=\frac{(-1)^{n+1}2^n}3+\frac{(-1)^{2n+2}}3\\
&=\frac{(-1)^{n+1}2^n+1}3\\
&=\frac{1-(-2)^n}3\;.
\end{align*}$$
This result can then be proved by mathematical induction, but I suspect that you’re not expected to go that far.
If you’ve already learned the summation formula for finite geometric series, you can apply it to get $s_n$ without looking at any patterns at all, and it’s something that you should learn as soon as possible if you don’t already know it. However, skill at pattern-recognition is useful anyway, so I thought that it might be useful to see how the problem can be attacked in that way as well.
In textbooks, the convergence of a sequences is often discussed before series.
If a sequence is monotonous, say $a_{n+1} \geq a_n$, and bounded from above, yes, it will converge. This is proven by showing it has an upper bound (reformulation of "bounded from above"), and using the fact that every bounded set has a smallest upper bound (called supremum). This smallest upper bound of the sequence is its limit.
Next, by considering the partial sums of a series, you will notice that the partial sums form a sequence. If this sequence of partial sums converges, the series is said to converge.
Your textbook seems to be confusing for you for not repeating the partial sum notation in the lines you quote. Just consider that writing the partial sum with a sum-sign is the shorthand for the many terms of the sum added together. The $s_n$ form the sequence of partial sums of the $a_n$.
If a sequence of partial sums $s_n$ converges, it follows that the sequence of its terms $a_n$ converge to 0.
Proof. For all n>N, $|s_n - s_{n+1}|$ will be arbitrarily small, hence $|a_{n+1}|$ will be arbitrarily small, i.e., the sequence of $a_n$ converges to 0.
Caution. The reverse is not true. If the sequence $(a_n)_{n \geq 1}$ converges to 0, then it does NOT follow that the sequence of its partial sums $(s_n)_{n \geq 1}$ converges.
Example. The sequence $(a_n)_n = (1/n)_n$ converges to 0, however the sequence of the partial sums $s_N = \sum_{n=1}^N a_n$ is unbounded (hence, the series diverges). Note that $\sum_{n=2^{k}+1}^{2^{k+1}} 1/n > 1/2$ for all k.
Best Answer
I can see two problems.
First off, your Fourier coefficients as given are only half as big as they should be to match your Series1 data. Perhaps you used the even nature of the target function so you only integrated over half the interval but then forgot to multiply by $2$ to take into account the half of the integral you didn't do.
Second, your value for $a_3$ is very high. I didn't actually look at your spreadsheet; here's my Matlab program:
And its output:
BTW, your expression for the Fourier series has a typo which I an gonna fix.