# Showing that the sum $\sum_{k=1}^n \frac1{k^2}$ is bounded by a constant

How can I shows that the summation of $1/k^2$ from k=1 to n is bounded above by a constant?

I could bind it by the geometric series from k=0 to n and add 1 to $(1/k^2)$ to get the ratio, r, and get $A_0 (1/(1-r))$. So is that going to be summation k=0 to n of $2/(k^2+1)$? And the ratio r, is $(2/(k^2+1))/(1/k^2)$?

You can't "bind" the series $\sum_{k=1}^{\infty}1/k^2$ with any geometric series because for any $r$ between $0$ and $1$ and any $a$, there will always be some $k$ such that $ar^k<1/k^2$ (that is, the ratio test fails for this series).
You can, however, say that (for $k>1$) $$\frac1{k^2}<\frac2{k(k+1)}= 2\left(\frac{1}{k}-\frac{1}{k+1}\right)$$
Now, what can we say about the sum $$\sum_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}\right)?$$