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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
$$sum_{i=0}^n a_ito S ,qquad sum_{i=0}^{n-1} a_ito S$$
as $ntoinfty$, so
$$a_n=Bigl(sum_{i=0}^n a_iBigr)-Bigl(sum_{i=0}^{n-1} a_iBigr)to S-S=0 .$$

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 $2epsilon<|a_{n+1}|$ 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.)

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, $a_i = left(-1right)^i$ 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.