Show that $\sum _{n=k }^{ l }\frac {1}{n^2}\le\frac { 1 }{k^2} +\frac {1}{k} -\frac {1}{l} $ holds

Question

I'm reviewing some Calculus 1. I want to show that

$$\sum _{n=k}^l\frac{1}{n^2}\le\frac{1}{k^2}+\frac{1}{k}-\frac{1}{l}$$

holds. I'm aware that I could make use of convergences and solve this problem by estimating some Integrals.

However, I want to use Calc 1 Methods, so basically I'm looking for some good estimates from above and below to proof the statement

Answer 1

\begin{eqnarray*} \sum_{n=k}^{l} \frac{1}{n^2} =\frac{1}{k^2} +\sum_{n=k+1}^{l} \frac{1}{n^2} \leq \frac{1}{k^2} +\sum_{n=k+1}^{l} \frac{1}{n(n-1)} =\frac{1}{k^2} +\sum_{n=k+1}^{l} \left(\frac{1}{n-1} - \frac{1}{n}\right) \\=\frac{1}{k^2} +\frac{1}{k} - \frac{1}{l}. \end{eqnarray*}

Answer 2

$$\sum_{n=k}^l\frac{1}{n^2}=\frac{1}{k^2}+\sum_{n=k+1}^l\frac{1}{n^2}\leq\frac{1}{k^2}+\sum_{n=k+1}^l\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{k^2}+\frac{1}{k}-\frac{1}{l}.$$