How to bound the tail of p-series

Question

How can I asses $S_n = \sum_{j=n}^\infty\frac{1}{j^p}, p>1$ in terms of $n$, specifically can I get something like $$S_n = O(?)$$

Answer 1

We can use the fact that for any decreasing function $f$ $$ \int_{s=a}^{b+1} f(s)\mathrm ds \le \sum_{i=a}^{b} f(i) \le \int_{s=a-1}^{b} f(s)\mathrm ds. $$ We obtain that $$ \frac{n^{1-p}}{p-1}\le\sum_{j=n}^\infty\frac1{j^{p}}\le\frac{(n-1)^{1-p}}{p-1}. $$ since $p>1$. Hence, $$ \sum_{j=n}^\infty\frac1{j^{p}}=O(n^{1-p}) $$ as $n\to\infty$.

Answer 2

$$\sum_{j=n}^\infty\frac{1}{j^p} = n^{-p+1} \times \frac 1n \sum_{j=n}^\infty\frac{n^p}{j^p} \sim n^{-p+1} \int_1^\infty \frac {du}{u^p} = n^{-p+1}\frac 1{p-1} $$