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.
You can use the extension to the real line and find the maximum by differentiation (of the logarithm, for convenience):
$$\left(\frac{\log x}x\right)'=\frac{1-\log x}{x^2}=0$$
Hence the function is decreasing on either sides of $x=e$ and the maximum for the discrete variable is one of $a_2, a_3$.
Best Answer
Consider the subsequences $b_{n_k},c_{n_j}$ which are respectively the even terms of $a_n$ and the odd terms of $a_n$. We have that $$\lim_{n_k \to \infty}b_{n_k}=\lim_{n_k \to \infty}\frac{2n_k^3}{n_k^3+1}=2$$ $$\lim_{n_j \to \infty}c_{n_j}=\lim_{n_j \to \infty}(-1)\frac{2n_j^3}{n_j^3+1}=-2$$ This excludes convergence of $a_n$.