Comparison Test

This tests utilises the ‘bounds’ of other Series, usually geometric or p-harmonic, to test a given series. It’s similar to the Sandwich Theorem, but doesn’t require two other series, since a series can either converge to a value or diverge (approach infinity).

Theorem §

Comparison Test

Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence of real numbers with positive terms only, i.e. $a_n\ge 0\forall n$ and ${\sum }_{n=1}^{\infty }{a}_n$ be it’s respective series.

Let $(b_n{)}_{n\in \mathbb{N}}$ be another sequence such that $\exists N\in \mathbb{N}$ such that $\forall k\in \mathbb{N},k>N$, $|a_k|\le b_k$. Meaning, after a certain point, every term in the sequence $a_n$ is lesser than $b_k$. Then:

$$\sum _{k=1}^{\infty }{b}_k\text{ converges}\to \sum _{k=1}^{\infty }{a}_k\text{ converges}\text{\hspace{1em}(1)}$$

Similarly, let $(c_n{)}_{n\in \mathbb{N}}$ be another sequence such that $\exists N\in \mathbb{N}$ such that $\forall k\in \mathbb{N},k>N$, $|a_k\ge c_k$. Meaning, after a certain point, every term in the sequence $a_n$ is greater than $c_k$. Then:

$$\sum _{k=1}^{\infty }{c}_k\text{ diverges}\to \sum _{k=1}^{\infty }{a}_k\text{ diverges}\text{\hspace{1em}(2)}$$

• It’s best to use geometric or harmonic series because their conditions for converging or diverging are simple.

https://www.desmos.com/calculator/8op6xpfuks

Proof §

Let $|s{|}_n={\sum }_{k=1}^n|a_k|$, the absolute partial sum. Let the number it converges to be $\beta$

Then:

$$\begin{array}{rl}|s{|}_n& =|a_1|+|a_2|+\cdots +|a_N|+\sum _{k=N+1}^n|a_k|\\ & \le |a_1|+|a_2|+\cdots +|a_N|+\sum _{k=N+1}^n{b}_k\\ & \le |a_1|+|a_2|+\cdots +|a_N|+\sum _{k=N+1}^{\infty }{b}_k\\ & =|a_1|+|a_2|+\cdots +|a_N|+\beta \end{array}\text{\hspace{1em}((bk)k∈N is monotone increasing)}$$

Hence it converges

Examples §

1: Divergence

Determine if the following series converges or diverges:

$$\sum _{n=1}^{\infty }\frac{n^2+4}{n^3+5}$$

We can see that, by looking at the fastest growing terms, the series approximately diverges: $\frac{n^2+4}{n^3+5}\sim \frac{n^2}{n^3}=\frac{1}{n}$ which is an example of a p-harmonic series. Now, we need to prove this using inequalities.
$\frac{n^2+4}{n^3+5}>\frac{n^2}{n^3+5}>\frac{n^2}{n^3+5n^3}$ (as $n\ge 0\forall n$) $\frac{n^2}{6n^3}=\frac{1}{6n}$ So, $\frac{1}{6n}\le \frac{n^2+4}{n^3+5}$ for every term of n, meaning we can use the comparison test.

$$\sum _{n=1}^{\infty }\frac{1}{6n}\text{diverges, as }\frac{1}{6}\text{ is a constant}$$

(using p-harmonic series, as well as the properties of divergent series)
Hence,

$$\sum _{n=1}^{\infty }\frac{n^2+4}{n^3+5}\text{diverges}$$

(Using comparison test)