convergence

harmonic series

${\sum }_{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\dots$

grouping by fractions that sum to $>\frac{1}{2}$
${\sum }_{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+\frac{1}{9}+\dots$
thus, doesn’t converge as sum would be $>\frac{1}{2}\cdot \infty$

if series is converted to form harmonic series, can prove that that series also doesn’t converge

ratio test

$\rho ={\lim }_{n\to \infty }|\frac{a_{n+1}}{a_n}|$
trying to make sure the limit ($\rho$) is $<1$ for convergence

comparison test

basically just the squeeze theorem
exists some $N$ s.t. $0\le a_n\le b_n$ for all $n\ge N$. if $\sum b_n$ converges, $\sum a_n$ converges
exists some $N$ s.t. $a_n\ge b_n\ge 0$ for all $n\ge N$. if $\sum b_n$ diverges, $\sum a_n$ diverges

with limits

if ${\lim }_{n\to \infty }\frac{a_n}{b_n}\ne 0$, then $\sum a_n$ and $\sum b_n$ both converge or both diverge
if ${\lim }_{n\to \infty }\frac{a_n}{b_n}=0$ and $\sum b_n$ converges, then $\sum a_n$ converges
if ${\lim }_{n\to \infty }\frac{a_n}{b_n}=\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges

alternating series

${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{b}_n=b_1-b_2+b_3-b_4$
or ${\sum }_n^{\infty }(-1{)}^{n+1}{b}_n=b_1-b_2+b_3-b_4$

converges if $0\le b_{n+1}\le b_n$ for all $n\le 1$
or if ${\lim }_{n\to \infty }{b}_n=0$

$S_N$ is sum from $1\to N$
$R_N$ is sum from $N+1\to \infty$
remainder $R_N=S-S_N$ satisfies $|R_N|\le b_{N+1}$
since $R_N$ becomes smaller when $-b_n+b_{n+1}$