How does one prove that $\sum\limits_{n=1}^{+\infty} \frac{1}{n^{1+2p}} \leq \frac{1+2p}{p}$?

Question

Got stuck on this while reading a paper. How does one prove that for $p>0$ the following inequality holds? $$\sum\limits_{n=1}^{+\infty} \frac{1}{n^{1+2p}} \leq \frac{1+2p}{p}$$ So far I got that the series' sum is the value of the Riemann $\zeta$-function at $z=1+2p$. But this gets me no further.

Answer 1

Note that since $x\to1/x^{1+2p}$ is decreasing in $[1,+\infty)$, $$\sum\limits_{n=1}^{+\infty} \frac{1}{n^{1+2p}}=1+\sum\limits_{n=2}^{+\infty} \frac{1}{n^{1+2p}}\leq 1+\sum\limits_{n=2}^{+\infty}\int_{n-1}^{n} \frac{dx}{x^{1+2p}}= 1+\int_{1}^{\infty}\frac{dx}{x^{1+2p}}=1+\frac{1}{2p}.$$

Answer 2

Using the Abel summation formula we have $$\sum_{n\leq N}\frac{1}{n^{1+2p}}=\frac{1}{N^{2p}}+\left(2p+1\right)\int_{1}^{N}\frac{\left\lfloor t\right\rfloor }{t^{2p+2}}dt$$ where $\left\lfloor t\right\rfloor $ is the floor function, so $$\sum_{n\geq1}\frac{1}{n^{1+2p}}=\left(2p+1\right)\int_{1}^{\infty}\frac{\left\lfloor t\right\rfloor }{t^{2p+2}}dt\leq\left(2p+1\right)\int_{1}^{\infty}\frac{1}{t^{2p+1}}dt=\frac{2p+1}{2p}.$$