Bounding a summation by an integral In progression of another question, my lecturer states $$\sum_{k=N+1}^\infty \frac{1}{k^2} \le \int^\infty_N \frac{1}{x^2}\, {\rm d}x.$$
However we have not covered bounding summations by integrals yet so I was wondering if someone could explain the choice of the function to integrate and why the bottom limit is $N$, rather than $N+1$ for example. Is this simply to ensure the integral is definitely greater?

If you look at the area under the curve for $1/k^2$ and the sum approximation it will become obvious why this equation makes sense:

The sum $\sum_{k=N+1}^\infty \frac{1}{k^2}$ are the bars and $\int_N^\infty \frac{1}{x^2}\mathrm{dx}$ is the area under the line.
The sum has to start 1 unit later, otherwise you would get this:

which actually shows that the sum starting at N will bound the integral from above.
You can combine these observations to form the equation:
$$ \sum_{k=N+1}^\infty \frac{1}{k^2} \le \int_{N}^\infty \frac{1}{x^2} \mathrm{dx} \le \sum_{k=N}^\infty \frac{1}{k^2} $$
