### Telescopic Series Telescopic series are very special and quite rare. They collapse in on themselves so that only the first term and the last term survive. This makes them easily summable. > [!EXAMPLE] > **a)** Show that $\sum^\infty_{n=1}\frac{1}{n^2+n}$ converges. > $$\begin{align} > &\frac{1}{n^2+n}\leq\frac{1}{n^2}\quad\text{and}\quad\sum^\infty_{n=1}\frac{1}{n^2} \\ > &\text{is a converging p-series }p=2>1\text{.} \\\\ > &\therefore\text{ By the comparison test, }\sum^\infty_{n=1}\frac{1}{n^2+n}\text{ also converges.} > \end{align}$$ > > **b)** By implementing a partial fraction decomposition of $\frac{1}{n^2+n}$ find the sum of the finite $N$th partial sum $S_n=\sum^N_{n=1}\frac{1}{n^2+n}$. This is called a **telescopic series**. > $$\begin{align} > &\frac{1}{n^2+n}=\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}=\frac{A(n+1)+B_n}{n(n+1)} \\\\ > &\therefore A(n+1)+B_n=1 \\\\ > &n=0:\quad A=1,\quad n=-1\Rightarrow-B=1\Rightarrow B=-1 \\\\ > &\therefore\frac{1}{n^2+n}=\frac{1}{n}-\frac{1}{n+1} \\\\ > &\sum^N_{n=1}\frac{1}{n^2+n}=\sum^N_{n=1}\frac{1}{n}-\frac{1}{n+1} \\\\ > S_N&=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{1}{3}-\frac{1}{4}+\cdots\frac{1}{N}-\frac{1}{N+1} \\ > &=1-\frac{1}{N+1} > \end{align}$$ > > **c)** Hence find the sum of the infinite series $\sum^\infty_{n=1}\frac{1}{n^2+n}$. > $$\begin{align} > S_\infty&=\sum^\infty_{n=1}\frac{1}{n^2+n}=\lim_{N\to\infty}1-\frac{1}{N+1}=1 > \end{align}$$ *** ### Alternating Series An alternating series takes the form $$\sum^\infty_{n=0}(-1)^na_n$$ where $a_n>0$. > [!EXAMPLE] > Write down the first 5 terms in the alternating series $$\sum^\infty_{n=4}(-1)^n\frac{1}{n^2}$$ > *** > $$\frac{1}{4^2}-\frac{1}{5^2}+\frac{1}{6^2}-\frac{1}{7^2}+\frac{1}{8^2}\cdots$$ #### The Leibnitz Test Consider the alternating series $$\sum^\infty_{n=0}(-1)^na_n$$where $a_n>0$. If $\sum^\infty_{n=0}a_n$ converges, then $\sum^\infty_{n=0}(-1)^na_n$ also converges. This is called **absolute convergence**. If $\sum^\infty_{n=0}a_n$ diverges, it is still possible for the original series to converge. This is called **conditional convergence** and is guaranteed when: 1. $a_{n+1} [!EXAMPLE] > Explore the convergence of the alternating series $$\sum^\infty_{n=4}(-1)^n\frac{1}{n^2}$$ > *** > **Absolute convergence** > Consider, $$\sum^\infty_{n=4}\frac{1}{n^2}$$ > This is a converging p-series $p=2>1$ > $\therefore$ By Leibnitz test $\sum^\infty_{n=4}(-1)^n\frac{1}{n^2}$ converges absolutely. > [!EXAMPLE] > Discuss the convergence of the alternating series $$\sum^\infty_{n=2}(-1)^n\frac{n}{n^2-1}$$ > *** > **Absolute convergence** > Consider, $$\sum^\infty_{n=2}\frac{n}{n^2-1}$$ > Now, $$\frac{n}{n^2-1}>\frac{n}{n^2}=\frac{1}{n}$$ > and $\sum^\infty_{n=2}\frac{1}{n}$ is a diverging harmonic series. > $\therefore$ No absolute convergence. > > **Conditional convergence** > Claim $\frac{n}{n^2-1}$ is decreasing. > > Consider, > $$\begin{align} > y&=\frac{x}{x^2-1} \\\\ > y\prime&=\frac{(x^2-1)(1)-x(2x)}{(x^2-1)^2} \\ > &=\frac{x^2-1-2x^2}{(x^2-1)^2} \\ > &=\frac{-1-x^2}{(x^2-1)^2}<0 \\\\ > &\therefore y\text{ is a decreasing function.} \\ > &\therefore a_{n+1} \end{align}$$ > Also, $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n}{n^2-1}=0$$ > $\therefore \sum^\infty_{n=2}(-1)^n\frac{n}{n^2-1}$ converges conditionally. > [!EXAMPLE] > Consider the alternating series, > $$\sum^\infty_{n=1}(-1)^{n+1}\frac{1}{n}=1-\frac{1}{1}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots$$ > *** > **a)** Prove that the series converges conditionally. > > **i)** Consider, $$\sum^\infty_{n=1}\frac{1}{n}$$ > This is a divergent p-series (harmonic series). > > **ii)** Let $a_n=\frac{1}{n}$ > $$\begin{align} > a_{n+1}&=\frac{1}{n+1}<\frac{1}{n}=a_n \\\\ > \lim_{n\to\infty}a_n&=\lim_{n\to\infty}\frac{1}{n}=0 > \end{align}$$ > > $\therefore \sum^\infty_{n=1}(-1)^{n+1}a_n$ converges conditionally. > >*** > > **b)** Earlier, we showed the Maclaurin series for $\ln(1+x)$ is $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\cdots$$ > Use this series to prove that $$1-\frac{1}{1}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots=\ln(2)$$ > Subbing in $x=1$, > $$\begin{align} > x=1\Rightarrow\ln(2)&=1-\frac{1^2}{2}+\frac{1^3}{3}-\frac{1^4}{4}\cdots \\ > &=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots > \end{align}$$ > > *** > > **c)** Suppose that the infinite series $$\ln(2)=1-\frac{1}{1}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots$$ is terminated at the $8$th partial sum $$S_8=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}$$ Find an error bound for the absolute difference between $S_8$ and $\ln(2)$. > > Just as in Maclaurin series, the error when terminating a convergent alternating series is bounded by the absolute value of the next term following the point of termination. > $\therefore\text{Error}\leq\frac{1}{9}$ > > *** > > **d)** How far does the series $$1-\frac{1}{1}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots$$ need to be taken in order for the difference between the series and $\ln(2)$ to be less than or equal to $0.001$? > $$\begin{align} > \frac{1}{n+1}&\leq\frac{1}{1000} \\\\ > \Rightarrow n+1&\geq1000 \\\\ > \Rightarrow n&\geq999 > \end{align}$$ > [!EXAMPLE] > Add up all the terms of $$1-\frac{1}{1}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots$$ in a different order to get a sum of $127$ instead of $\ln(2)$. > *** > $$\begin{align} > &\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right)\text{ until}>127 \\\\ > &\left(-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\cdots\right)\text{ until}<127 \\\\ > &\left(\text{Select from remainder of V}\right)\text{ until}>127 \\\\ > &\left(\text{Select from remainder of U}\right)\text{ until}<127 > \end{align}$$ > Final sum will converge to $127$. *** ### Power Series A series of the form $\sum^\infty_{n=0}a_nx^n$ is called a power series in powers of $x$. A series of the form $\sum^\infty_{n=0}a_n(x-\beta)^n$ is called a power series in powers of $(x-\beta)$. For convergence, the ratio test, $$\lim_{n\to\infty}\left|\frac{T_{n+1}}{T_n}\right|<1$$ > [!EXAMPLE] > Determine an open interval $I$ of convergence and the radius $R$ of convergence for the power series, $$\sum^\infty_{n=0}\frac{(x-2)^n}{n+5}$$ > *** > $$\text{Let}\quad T_n=\frac{(x-2)^n}{n+5}\quad\text{and}\quad T_{n+1}=\frac{(x-2)^{n+1}}{n+6}\text{.}$$ > Then, > $$\begin{align} > \lim_{n\to\infty}\left|\frac{T_{n+1}}{T_n}\right|&= > \lim_{n\to\infty}\left|\frac{\frac{(x-2)^{n+1}}{n+6}}{\frac{(x-2)^n}{n+5}}\right|\\ > &=\lim_{n\to\infty}\left|\frac{(x-2)^{n+1}}{n+6}\times\frac{n+5}{(x-2)^n}\right|\\ > &=\lim_{n\to\infty}\left|(x-2)\times\frac{n+5}{n+6}\right|\\ > &=\left|x-2\right|\lim_{n\to\infty}\left(\frac{n+5}{n+6}\right)\\ > &=\left|x-2\right|<1\\\\ > &x-2<1\quad\text{OR}\quad x-2>-1\\ > &x<3\quad\text{OR}\quad x>1\\ > &1 &I=(1,3)\\ > &R=1 > \end{align}$$ > This means that the power series $\sum^\infty_{n=0}\frac{(x-2)^n}{n+5}$ only converges absolutely for $1 [!EXAMPLE] > Determine an open interval $I$ of convergence and the radius $R$ of convergence for the power series, $$\sum^\infty_{k=0}\frac{(-5)^k(x-3)^k}{k^2+1}$$ > *** > $$\begin{align} > \text{Let}\quad T_k=\frac{(-5)^k(x-3)^k}{k^2+1}\quad\text{and}\quad T_{k+1}&=\frac{(-5)^{k+1}(x-3)^{k+1}}{(k+1)^2+1}\\\\ > &=\frac{(-5)^{k+1}(x-3)^{k+1}}{k^2+2k+2} > \end{align}$$ > Then, > $$\begin{align} > \lim_{k\to\infty}\left|\frac{T_{k+1}}{T_k}\right|&= > \lim_{k\to\infty}\left|\frac{(-5)^{k+1}(x-3)^{k+1}}{k^2+2k+2}\times\frac{k^2+1}{(-5)^k(x-3)^k}\right|\\ > &=\lim_{k\to\infty}\left|\frac{-5(x-3)(k^2+1)}{k^2+2k+2}\right|\\ > &=5\left|x-3\right|\lim_{k\to\infty}\left(\frac{k^2+1}{k^2+2k+2}\right)\\ > &=5\left|x-3\right|\times1<1\\ > &=\left|x-3\right|<\frac{1}{5}\\\\ > &x-3<\frac{1}{5}\quad\text{OR}\quad x-3>-\frac{1}{5}\\ > &x<\frac{16}{5}\quad\text{OR}\quad x>\frac{14}{5}\\ > &\frac{14}{5} &I=\left(\frac{14}{5},\frac{16}{5}\right)\\ > &R=\frac{1}{5} > \end{align}$$ > This means that the power series $\sum^\infty_{k=0}\frac{(-5)^k(x-3)^k}{k^2+1}$ only converges absolutely for $\frac{14}{5} [!EXAMPLE] > Determine an open interval $I$ of convergence and the radius $R$ of convergence for the Maclaurin series, $$e^x=\sum^\infty_{k=0}\frac{x^k}{k!}$$ > *** > $$\text{Let}\quad T_k=\frac{x^k}{k!}\quad\text{and}\quad T_{k+1}=\frac{x^{k+1}}{(k+1)!}$$ > Then, > $$\begin{align} > \lim_{k\to\infty}\left|\frac{T_{k+1}}{T_k}\right|&= > \lim_{k\to\infty}\left|\frac{x^{k+1}}{(k+1)!}\times\frac{k!}{x^k}\right|\\ > &=\lim_{k\to\infty}\left|k\times\frac{1}{k+1}\right|\\ > &=\left|x\right|\lim_{k\to\infty}\left(\frac{1}{k+1}\right)\\ > &=\left|x\right|\times0<1\\ > &\text{True for all }x\text{.}\\\\ > &I=\mathbb{R}\\ > &R=\infty > \end{align}$$ > The Maclaurin series for $e^x$ converges for all real $x$ and hence the radius of convergence is infinite.

MATH1231