Showing that $\sum_{k\mathop=n}^{\infty} \frac 1 {k^2} < \frac 2 {2n-1}$ I need to show that

$$\sum_{k\mathop=n}^{\infty} \frac 1 {k^2} &lt; \frac 2 {2n-1}$$

Could I have a small hint? 

I have tried many things, like bounding it above with the corresponding integral but this is too weak. I have tried subtracting $1$ from each of the denominators (excluding the first term) and then evaluating the resulting telescoping sum, but even that turns out to be too weak.
$$\frac1{k^2}&lt;\frac1{k^2-\frac14}=\frac1{k-\frac12}-\frac1{k+\frac12},$$ so
$$\sum^\infty_{k=n}\frac1{k^2}&lt;\sum^\infty_{k=n}\left(\frac1{k-\frac12}-\frac1{k+\frac12}\right)=\frac1{n-\frac12}=\frac2{2n-1}$$
Hint: Compare with the integral of $1/x^2$, by comparison with the integral, you obtain that $\sum_{k\geq n}{1\over k^2}&lt;\int_{n-{1\over 2}}^{+\infty}{1\over x^2}dx={2\over{2n-1}}$.