- Geometric Serise
$ \sum_{n=0}^\infty a r^{n} = \frac{a}{1-r} = \frac{ \text{leading term}}{1-\text{ratio}} , \hspace{1em} |r|<1.$
- limit of rational type
If order of $P$ = order of $Q$, $\hspace{1em} \lim_{n\to\infty} \frac{P(n)}{Q(n)} = \frac{\text{highest order coefficient of }P(n)}{\text{highest order coefficient of }Q(n)} $
$$e.g., \hspace{2em} \lim_{n\to\infty}\frac{(5n+1)^3+\sqrt{1+16n^6}}{9+7n+5n^3} = \frac{5^3+\sqrt{16}}{5} = \frac{129}{5}, \hspace{1em} \text{order} = 3 $$
- L'Hôpital's rule is handy sometimes.
- Convergence and divergence
$$\fcolorbox{white}{yellow}{ Partial Sum: $\hspace{1em} S_n = a_1+a_2+\cdots +a_n = \sum_{j=1}^n a_j $} \hspace{1em}
\fcolorbox{white}{yellow}{ Series: $\hspace{1em} S = \lim_{n\to\infty} S_n = \sum_{j=1}^\infty a_j $ } \hspace{1em}
\fcolorbox{white}{yellow}{ Remainder: $\hspace{1em} R_n= S-S_n \rule[-3ex]{0pt}{6.5ex} $} $$
- Preliminary test (Divergence test):
If $\hspace{1em} \lim_{n\to\infty} a_n \neq 0$, then $ \sum_{n=1}^\infty a_n $ is divergent