I came across this bound in a lecture, and I don't know how to prove it or it's generalization.
$$\sum_{k\geq n} \frac{1}{k^2}\leq \frac{c}{n}$$ for some constant $c$.
How do you prove the above? and how do you generalize this inequality for $0<p<2$ and $$\sum_{k\geq n} \frac{1}{k^p}$$
Thank you.
First notice that $\frac{1}{k^2}\leqslant\int_{k-1}^k\frac{dx}{x^2}$, then for $n\geqslant 2$ $$ \sum_{k\geqslant n}\frac{1}{k^2}\leqslant\sum_{k\geqslant n}\int_{k-1}^k\frac{dx}{x^2}=\int_{n-1}^{+\infty}\frac{dx}{x^2}=\frac{1}{n-1}\leqslant\frac{2}{n} $$ You can do the same to get for $p>1$, $$ \sum_{k\geqslant n}\frac{1}{k^p}\leqslant\frac{2^p}{n^p} $$