# Calc6.8

Discuss the convergence of $\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{p}}}$.

This type of series is called a p-series. The idea of this problem is to find which values of p give us a convergent series and which values of p give us a divergent series. When we have that info, we can find the convergence or divergence of any p-series without having to calculate hundreds of different integrals separately, simply by knowing the value of p. Use the integral test to determine convergence and divergence.

$\int _{{1}}^{{\infty }}{\frac {1}{x^{p}}}={\frac {x^{{-p+1}}}{-p+1}}{\bigg |}_{{1}}^{{\infty }}={\frac {1}{1-p}}{\frac {1}{x^{{p-1}}}}{\bigg |}_{{1}}^{{\infty }}$

First, if $p=1$ the integral yields

$\ln |x|{\bigg |}_{{1}}^{{\infty }}=\lim _{{b\to \infty }}\lim _{{a\to 1^{+}}}\left(\ln b-\ln a\right)=\lim _{{b\to \infty }}\lim _{{a\to 1^{+}}}\ln {\frac {b}{a}}=\infty \,$

So, when $p=1$ the integral diverges.

If $p>1$ the integral converges because as $x\rightarrow \infty$ the integral goes to $0$.

If $p<1$ the integral diverges because as $x\rightarrow \infty$ the integral goes to $\infty$.

So, by the integral test, the series

$\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{p}}}\,$

converges when $p>1$ and diverges otherwise, for any real p.