Convergence Tests for Series of Positive Real Numbers

//This page is intended for the Complex Analysis hub and is only a brief overview of the many convergence tests for positive series of real numbers. A more in-depth look at these topics can be found on the Real Analysis page.

# Convergence Tests for Series of Positive Real Numbers

We will now state some of the common convergence tests for series of positive real numbers and provide an example of applying each of them. Note that some of these tests actually apply to nonpositive series!

 Theorem 1 (The Divergence Test): Consider the series $\displaystyle{\sum_{n=0}^{\infty} a_n}$. If $\displaystyle{\lim_{n \to \infty} a_n \neq 0}$ then $\displaystyle{\sum_{n=0}^{\infty} a_n}$ diverges.

For example, the series $\displaystyle{\sum_{n=1}^{\infty} n}$ diverges since $\displaystyle{\lim_{n \to \infty} n = \infty \neq 0}$.

 Theorem 2 (The Geometric Series Test): Consider the series $\displaystyle{\sum_{n=0}^{\infty} r^n}$. a) If $\mid r \mid < 1$ then this series converges to $\displaystyle{\frac{1}{1 - r}}$. b) If $\mid r \mid \geq 1$ then this series diverges.

For example, the series $\displaystyle{\sum_{n=0}^{\infty} \frac{1}{3^n} = \sum_{n=0}^{\infty} \left ( \frac{1}{3} \right )^n}$ converges since $\displaystyle{\mid r \mid = \biggr \lvert \frac{1}{3} \biggr \rvert < 1}$ and the sum of the series is $\displaystyle{\frac{1}{1 - \frac{1}{3}} = \frac{3}{2}}$.

 Theorem 3 (The p-Series Test): Consider the series $\displaystyle{\sum_{n=1}^{\infty} n^p}$. a) If $p \geq -1$ then this series diverges to $\infty$. b) If $p < -1$ then this series converges.

Consider the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty} n^{-2}}$. We have that $p = -2 < -1$ so this series converges.

 Theorem 4 (The Comparison Test): Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ be two positive series of real numbers such that $a_n \leq b_n$ for all $n \in \mathbb{N}$. a) If $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges. b) If $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges then $\displaystyle{\sum_{n=1}^{\infty} b_n}$ diverges.

For example, consider the series $\displaystyle{\sum_{n=1}^{\infty} \frac{n+1}{n}}$. For all $n \in \mathbb{N}$ we have that $\displaystyle{\frac{1}{n} \leq \frac{n+1}{n}}$. But the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n}}$ diverges by the p-series test and so $\displaystyle{\sum_{n=1}^{\infty} \frac{n+1}{n}}$ diverges. The divergence of this series can also be obtained by applying the divergence test.

 Theorem 5 (The Ratio Test): Consider the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and let $\displaystyle{L = \lim_{n \to \infty} \biggr \lvert \frac{a_{n+1}}{a_n} \biggr \rvert}$. Provided that $L$ exists: a) If $L < 1$ then this series converges absolutely. b) If $L > 1$ then this series diverges. c) If $L = 1$ then this test is inconclusive.

For example, consider the series $\displaystyle{\sum_{n=1}^{\infty} \frac{n^2}{n!}}$. Then by applying the ratio test we have that:

(1)
\begin{align} \quad L = \lim_{n \to \infty} \biggr \lvert \frac{(n+1)^2}{(n+1)!} \cdot \frac{n!}{n^2} \biggr \rvert = \lim_{n \to \infty} \frac{(n+1)^2}{(n+1)n^2} = \lim_{n \to \infty} \frac{n+1}{n^2} = 0 < 1 \end{align}

Therefore the series $\displaystyle{\sum_{n=1}^{\infty} \frac{n^2}{n!}}$ converges by the ratio test.

 Theorem 6 (The Root Test): Consider the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and let $\displaystyle{L = \lim_{n \to \infty} (\mid a_n \mid)^{1/n}}$. Provided that $L$ exists: a) If $L < 1$ then this series converges absolutely. b) If $L > 1$ then this series diverges. c) If $L = 1$ then this test is inconclusive.

For example, consider the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^n}}$. Then by applying the root test we have that:

(2)
\begin{align} \quad L = \lim_{n \to \infty} \biggr \lvert \frac{1}{n^n} \biggr \rvert^{1/n} = \lim_{n \to \infty} \frac{1^{1/n}}{n} = \lim_{n \to \infty} \frac{1}{n} = 0 < 1 \end{align}

Therefore $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^n}}$ converges by the root test.