$$\zeta(3)=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}H_n^{\phi(2)}(2n+1)}{n(n+1)}$$ Where, $H_n^{\phi(x)}$, is the n-th Harmonic number of the alternating-zeta series $\phi(x)$. Here in this case;
$$H_n^{\phi(2)}=1-\frac{1}{2^2}+\frac{1}{3^2}-...\pm\frac{1}{n^2}$$
To actually put what I commented into an answer, probably the best way to see why this sum converges, is to expand out the summands for the Harmonic numbers in rows: $$\frac {3}{2}(1)$$ $$-\frac {5}{6}(1-\frac {1}{4})$$ $$+\frac {7}{12}(1-\frac{1}{4}+\frac{1}{9})$$ ... Now, instead of computing the sum in rows, we look at the sums of the columns.in the column of ones, we find $$1(\frac{3}{2}-\frac {5}{6}+\frac {7}{12}...\frac {2n+1}{n(n+1)}...)$$ A little tinkering with these numbers reveals that this is in fact equal to $$1((1+\frac{1}{2})-(\frac {1}{2}+\frac {1}{3})+(\frac {1}{3} + \frac{1}{4})...(\frac {1}{n} + \frac {1}{n+1})...)$$ The nth partial sum here is $1+\frac {1}{n+1}$, which tends to one as n goes to infinity, so the coefficient of one in this sum. When we expand the column with fourths, a similar analysis reveals that the total coefficient on one fourth is one half. And in general the coefficient on $\frac 1{n^2}$ is 1/n. So adding all these together, the summands are $\frac {1}{n^3} $, giving the normal sum for $\zeta (3)$.