I can prove this sum has a constant upper bound like this:
$$\sum_{k=1}^n \frac1{k ^ 2} \lt 1 + \sum_{k=2}^n \frac 1 {k (k - 1)} = 2 - \frac 1 n \lt 2$$
And computer calculation shows that sum seems to converge to 1.6449. But I still want to know:
A sequence that is increasing and bounded must converge. That's one of the fundamental properties of the real line. So once you've observed that your sequence of partial sums is bounded, since it obviously increases, it must converge. Of course it is a very famous series, and it converges to a number which quite miraculously has a "closed form" formula: it is $\pi^2/6$.
EDIT: for many proofs of this famous formula, see this MO question.
Yes, it does converge, and believe it or not, $$\sum_{k=1}^\infty\frac{1}{k^2}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$$ Determining the specific value of this infinite sum was originally known as the Basel problem, and Euler was the first person to determine the correct value of the sum, although his initial methods were not 100% rigorous (but they can be made to be rigorous, and zyx points out below that Euler later gave valid proofs).
In general, the infinite sum $$\sum_{k=1}^\infty\frac{1}{k^s}$$ is extremely important in number theory, so much so that we give it a name, the Riemann zeta function, $\zeta(s)$. That is, $$\zeta(s)=\sum_{k=1}^\infty\frac{1}{k^s}.$$ So, we see that $\zeta(2)=\frac{\pi^2}{6}$. Wikipedia lists some other known values of the zeta function. A priori, the function $\zeta(s)$ is defined for any real number $s>1$ (that is, the sum $\sum_{k=1}^\infty\frac{1}{k^s}$ will converge for any $s>1$); but in fact, there is a well-defined notion of raising integers to complex numbers, and then we get that $\zeta(s)$ is well-defined for every $s\in\mathbb{C}$ for which $\text{Re}(s)>1$. Using something called analytic continuation, we can then define $\zeta(s)$ for every complex number (other than $s=1$).
