How do we bound the function $\frac{-1}{2}\sum_{m=n}^{\infty}\frac{1}{m^{2}}$? In case the question didn't display in the title correctly: How do we bound the function $\frac{-1}{2}\sum_{m=n}^{\infty}\frac{1}{m^{2}}$?
I think a way I can do this is to show that $$\sum_{m=n}^{\infty}\frac{1}{m^{2}} < \frac{1}{n^{2}} + \int_{n}^{\infty}\frac{1}{x^{2}}$$
But I'm a little unsure of the process it takes to get there.
 A: Estimating via the integral works well. Another way is to note that the tail (if we forget about the $-\frac{1}{2}$ in front) is less than
$$\frac{1}{(n-1)(n)}+\frac{1}{(n)(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots.$$
But $\frac{1}{(n-1)(n)}=\frac{1}{n-1}-\frac{1}{n}$ and the next term is $\frac{1}{n}-\frac{1}{n+1}$, and the next term is $\frac{1}{n+1}-\frac{1}{n+2}$, and so on.  Now note the massive cancellation (telescoping). We get $\frac{1}{n-1}$.
We conclude that the tail is less than $\frac{1}{n-1}$.  A similar argument shows that the tail is greater than $\frac{1}{n}$, so we have obtained reasonably tight bounds for the tail.
Remark: Since the post expresses uncertainty about the estimation with the integral, we give a brief explanation. The sum $\sum_{n+1}^\infty \frac{1}{k^2}$ can be thought of as the combined area of a bunch of rectangles, all of base $1$. The base of the first rectangle is $[n,n+1]$ and its height is $\frac{1}{(n+1)^2}$. The base of the second rectangle is $[n+1,n+2]$ and its height is $\frac{1}{(n+2)^2}$, and so on.
Draw the curve $y=\frac{1}{x^2}$. In the interval $[n,n+1]$ this curve is above the first rectangle, so the first rectangle has area less than $\int_n^{n+1}\frac{dx}{x^2}$. In the same way, the second rectangle has area less than $\int_{n+1}^{n+2}\frac{dx}{x^2}$, and so on. So the sum of the areas of all the rectangles is less than $\int_n^\infty \frac{dx}{x^2}$. 
A: The process is this: For every $m>n$ and for $m-1 \leq x <m$ let $f(x)=1/m$. And let $g(x)=1/x$ for all $x >0$. Then for all $m>n$ we have  $f(x) \le g(x)$ with equality only when $x$ is an integer greater than $n $.Therefore $$\sum_{m=n}^{\infty} 1/m^2=1/n^2 + \sum_{m=n+1}^{\infty} 1/m^2=$$ $$=1/n^2 +\sum_{m=n+1}^{\infty} \int_{x=m-1}^{x=m} f(x) dx =$$= $$=1/n^2+\int_{x=n}^{\infty} f(x) dx $$ $$ <1/n^2+\int_{x=n}^{\infty} g(x) dx =$$ $$= 1/n^2 +\int_{x=n}^{\infty}( 1/x) dx=1/n^2 +1/n.$$
