Suppose I have a sequence $\{a_i\}_{i=1}^{\infty}$. I know that $a_i \geq 0$ for all $i$ and $\sum_{i=1}^{\infty}a_i < \infty$. Can I say that
$$ \lim_{i \to \infty} a_i = 0$$
Intuitively, I think this is true. If the limit tends to a positive number, the sum would explode as well. However, I'm having a hard time proving it formally. Can I get a hint?
We have $$\sum_{i=0}^n a_i\to S\ ,\qquad \sum_{i=0}^{n-1} a_i\to S$$ as $n\to\infty$, so $$a_n=\Bigl(\sum_{i=0}^n a_i\Bigr)-\Bigl(\sum_{i=0}^{n-1} a_i\Bigr)\to S-S=0\ .$$
I'm not quite sure if the sequence starts at index $0$ or $1$ as your question uses $1$ as the sequence start but the summation starts at $0$. As it's not important, I will assume it's $0$. Note that
$$\sum_{i \, = \, 0}^{\infty} a_i \lt \infty \tag{1}\label{eq1}$$
usually means it converges to a limit. However, consider the somewhat looser restriction of there just simply existing a real supremum, call it $S$, so that
$$\sum_{i \, = \, 0}^{n} a_i \le S \; \forall \; n \ge 0 \tag{2}\label{eq2}$$
As for trying to show the limit of $a_i$ is $0$, I will just let the "$- \; 0$" part be implied. Also, it's given that $a_i \ge 0$, so the partial sums are non-decreasing, and there are no non-negative values so I don't need to use absolute values in the following. For proving that
$$\lim_{i \, \to \, \infty} a_i = 0 \tag{3}\label{eq3}$$
consider that for any $\epsilon \gt 0$ there must be a finite number of integers for which $a_n \ge \epsilon$ as, otherwise, the sum of them would be unbounded. Thus, there must be either none or a maximum index, but in either choose an integer $n_0$ to be greater than any such value and, thus, $a_n \lt \epsilon$ for all $n \gt n_0$, which confirms that \eqref{eq3} is true. I won't show it here, but the infinite sum has a limit as well of $S$.
Note that assuming just a supremum for the sums without assuming that $a_i \ge 0$ won't suffice to show the limit as, for example, $a_i = \left(-1\right)^i$ would cause the partial sums to alternate between $1$ and $0$, thus allowing a supremum of $1$, but with $a_i$ not converging to $0$.
Note that requiring the sum of $a_i$ to have a finite limit means that, as David indicated in his answer's comment, also requiring the $a_i$ to be non-negative is unnecessary. However, if the requirement of \eqref{eq1} includes the possibility of the limit being $-\infty$, then some appropriate restrictions on $a_i$ would be required.
Informal outline for a more formal proof:
Let $s_n$ be the sum of the first $n$ terms and let the limit of the series be $m$.
Convergence means that for as small an $\epsilon>0$ as we like and large enough $N$, $n>N$ guarantees $|m-s_n|<\epsilon$.
But this goes awry if $\{a_n\}$ doesn't converge to $0$, because when $2\epsilon<|a_{n+1}|$ it's not possible to have both $|m-s_n|<\epsilon$ and $|m-s_{n+1}|<\epsilon$.
(And use the definition of $\{a_n\}$ not converging to show that this situation must arise.)
