Calc6.8

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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.


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