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?

Note that $\dfrac1{x^2}$ is a decreasing function on (a,b), where $a>0$. Hence, on $(a,b)$, we have
$$\dfrac1{b^2} \leq \dfrac1{x^2} \leq \dfrac1{a^2}$$
Hence, on a unit interval $(m-1,m)$, we have
$$\dfrac1{m^2} \leq \dfrac1{x^2}$$
We hence have
$$\sum_{m=N+1}^{\infty} \dfrac1{m^2} \leq \int_N^{\infty} \dfrac{dx}{x^2}$$
