% This is lecture no. 25 of MH1101/: Calculus II.

\chapter{Finding Limits using Power Series}%
\label{cha:finding_limits_using_power_series}

\section{Using known Power Series}%
\label{sec:using_known_power_series}

We can use power series to evaluate the limit of a function.
Here are some important Maclaurin series:
\begin{center}
	\begin{tabular}{c|c}
		\hline
		$\ds{\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}=1+x+x^2+x^3+\dotsb}$ & $R=1$\\
		$\ds{e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb}$ & $R=\infty$\\
		$\ds{\sin x = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\dotsb}$ & $R=\infty$ \\
		$\ds{\cos x = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n}}{(2n!)}=1-\frac{x^2}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\dotsb}$ & $R=\infty$ \\
		$\ds{\tan^{-1}x = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\dotsb}$ & $R=1$ \\
		$\ds{\ln(1+x)=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^{n}}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^{4}}{4}+\dotsb}$ & $R=1$ \\
		$\ds{(1+x)^{k}=\sum_{n=0}^{\infty} \binom{k}{n}x^{n} = 1 + kx+\frac{k(k-1)}{2!}x^2 + \dotsb}$ & $R=1$\\
		\hline
	\end{tabular}
\end{center}
\begin{exmp}
	Find the sum of the series:
	\[
	\frac{1}{1\cdot 2}-\frac{1}{2\cdot 2^2} + \frac{1}{3\cdot 2^3} - \frac{1}{4\cdot 2^{4}}+\dotsb
	\]
	\begin{sol}
		Using sigma notation, we can write the series as:
		\[
			\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n\cdot 2^{n}} = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\left(\frac{1}{2}\right)^{n}}{n}
		.\]
		From the table above, this series matches the entry for $\ln(1+x)$ with $x=\frac{1}{2}$.
		So:
		\[
			\sum_{n=0}^{\infty} (-1)^{n-1}\frac{1}{n\cdot 2^{n}}=\ln\left(1+\frac{1}{2}\right) = \ln \frac{3}{2}
		.\]
	\end{sol}
\end{exmp}
