Comparison Theorem for Series

Theorem

Divergent Case

1. Given ${\sum }_{n=1}^{\infty }{a}_n$, ${\sum }_{n=1}^{\infty }{c}_n$,
2. If $\forall n\in \mathbb{N}$, $0\le c_n\le a_n$
3. And ${\sum }_{n=1}^{\infty }{c}_n$ Diverges Then, ${\sum }_{n=1}^{\infty }{a}_n$ also diverges

Convergent Case

1. Given ${\sum }_{n=1}^{\infty }{a}_n$, ${\sum }_{n=1}^{\infty }{b}_n$,
2. If $\forall n\in \mathbb{N}$, $0\le a_n\le b_n$
3. If ${\sum }_{n=1}^{\infty }{b}_n$ Converges Then, ${\sum }_{n=1}^{\infty }{a}_n$ also converges

Proof

Proof for Convergent Case

1. Suppose ${\sum }_{n=1}^{\infty }{a}_n$, ${\sum }_{n=1}^{\infty }{b}_n$, exists
2. Suppose $0<a_n<b_n$
3. Let $n\in \mathbb{N}$ be arbitrary
4. Consider $S_{n+1}=S_n+a_{n+1}$ by defn of Series Partial Sum
5. Note that $S_n\ge 0,a_{n+1}\ge 0$
6. Thus, $S_{n+1}>S_n$, which means $\{S_n\}$ is increasing
7. Since, $\forall n\in \mathbb{N},a_n\le b_n$
8. $⟹{\sum }_{n=1}^{\infty }{a}_n\le {\sum }_{n=1}^{\infty }{b}_n$
9. But, since $S_n\le {\sum }_{n=1}^{\infty }{a}_n$ as $n\in \mathbb{N}⟹n<\infty$ by defn of infinite, as $\{S_n\}$ is increasing
10. So, $S_n\le {\sum }_{n=1}^{\infty }{b}_n\in \mathbb{R}$ as ${\sum }_{n=1}^{\infty }{b}_n$ converges
11. By BMCT, $\{S_n\}$ converges
12. Therefore, ${\sum }_{n=1}^{\infty }{a}_n$ converges by definition

Example

Find if ${\sum }_{n=1}^{\infty }\frac{\arctan (n)}{n^{0.5}+4^n}$ converges or diverges

Soln

Proof

1. For $n\in \mathbb{N}$
2. $a_n=\frac{\arctan (n)}{n^{0.5}+4^n}>0$
3. Note that $4^n$ is increasing at a fastest rate
4. Note that $\arctan (n)<\frac{\pi }{2}$
5. Then, $a_n<\frac{\pi }{2}\left(\frac{1}{n^{0.5}+4^n}\right)$
6. $<\frac{\pi }{2}\left(\frac{1}{4^n}\right)$
7. By GS Test, our series does indeed convger