Introduction

Welcome to this guide on p-Series and Harmonic series, this article will guide you through the following:

p-Series

A p-series is a series that has the form of

$\sum _{n=1}^{\infty }\frac{1}{n^p}=1+\frac{1}{2^p}+\frac{1}{3^p}+...+\frac{1}{n^p}$

Where $p>0$.

Under what condition does a p-series converge?

Through integral test, one can determine the convergence of a p-series. First, check if the requirements of the integral test are met:

This means we can apply the integral test to this series, so set up this integral:

${\int }_1^{\infty }\frac{1}{x^p}dx$

$=\underset{b\to \infty }{\lim }{\int }_1^b\frac{1}{x^p}dx$

$=\underset{b\to \infty }{\lim }\frac{1}{-p+1}{x}^{-p+1}{|}_1^b$

$=\underset{b\to \infty }{\lim }\frac{1}{-p+1}{b}^{-p+1}-\frac{1}{-p+1}$

We know that $\frac{1}{-p+1}$ is a real number that has a finite value, so we should examine this limit $\underset{b\to \infty }{\lim }\frac{1}{-p+1}{b}^{-p+1}$ to determine the convergence of the series.

$\underset{b\to \infty }{\lim }\frac{1}{-p+1}{b}^{-p+1}$

$=\frac{1}{-p+1}\underset{b\to \infty }{\lim }{b}^{-p+1}$

This limit has a finite value when $-p+1<0$, or $p>1$. Thus the series converges if $p>1$, and diverges if $0<p\le 1$.

When $p=1$, the series is called the harmonic series, which diverges.

Note that if $p<0$, say $p=-q$, where $q$ is a positive number for instance, then we can rewrite the series as being:

$\sum _{n=1}^{\infty }\frac{1}{n^{-q}}=\sum _{n=1}^{\infty }{n}^q$

When running the $n$th term test upon this series, we find:

$\underset{n\to \infty }{\lim }{n}^q=\infty$

Which thus means the original series diverges (as $q$ is positive).

Example 1:

$\sum _{n=1}^{\infty }\frac{1}{\sqrt[4]{k^3}}$

The exponent can be written as $\frac{3}{4}<1$, which means it diverges.

Practice

Determine if the following series converge: