How to prove $\lim _{n \to \infty} S_n=\infty$ for partial sums of a divergent series? If $\sum_{n=1}^\infty a_n $ diverges and $a_n\geq 0 $, then
$\lim _{n \to \infty} S_n=\lim_{n \to \infty} \sum_{k=1}^n a_k =\infty $
Sol :
If $\sum_{n=1}^\infty a_n $diverges
Then,
$ \lim_{n \to \infty}a_n \not=0$
I don't know how complete  it, any hint or note for prove this problem .
 A: $a_n\ge 0$, then $(S_n)_{n\in\mathbb N}=\sum_{k=1}^n a_k$ is a increasing sequence.
Therefore, $\lim_nS_n\ge0$ (possibly $\infty$)
If $\lim_nS_n\lt\infty$, then by definition of series, the series $\sum_{k=1}^\infty a_k$ is said to be convergent, i.e., doesn't diverges.
Also, the solution you gave is wrong, as you can consider $\sum_{k=1}^\infty\frac{1}{k}=+\infty$ but $\lim_k\frac{1}{k}=0$.
A: The non-negativity of the sequence $(a_n)$ implies that $(S_n)$ is a monotonically increasing sequence since $S_{n+1}-S_n = a_{n+1} \geq 0$ for each $n$.
So, monotone convergence implies that there exists some extended real number (extended real number being either a real number or infinity) $x$ such that $(S_n)$ converges to $x$. 
If $x \in [0, \infty)$ then $(S_n)$ is a convergent sequence and the series is convergent contrary to assumption. So, $x$ must be the extended real number $\infty$ which is to say that $(S_n)$ “diverges to infinity.” 
Your book might not consider extended real numbers with the theorem on monotone convergence but adjusting for this shouldn’t be difficult.
