series 1 lecture participation: problem 2 (2 points) calculate $s_3$, $s_4$ and $s_5$ and then find the sum…

series 1 lecture participation: problem 2 (2 points) calculate $s_3$, $s_4$ and $s_5$ and then find the sum for the telescoping series $s=sum_{n = 4}^{infty}(\frac{1}{n + 1}-\frac{1}{n + 2})$ where $s_k$ is the partial sum using the first $k$ values of $n$. $s_3$: blank $s_4$: blank $s_5$: blank $s$: blank note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

series 1 lecture participation: problem 2 (2 points) calculate $s_3$, $s_4$ and $s_5$ and then find the sum for the telescoping series $s=sum_{n = 4}^{infty}(\frac{1}{n + 1}-\frac{1}{n + 2})$ where $s_k$ is the partial sum using the first $k$ values of $n$. $s_3$: blank $s_4$: blank $s_5$: blank $s$: blank note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Answer

Explanation:

Step1: Calculate (S_3)

[ \begin{align*} S_3&=\sum_{n = 4}^{6}\left(\frac{1}{n + 1}-\frac{1}{n+2}\right)\ &=\left(\frac{1}{4 + 1}-\frac{1}{4+2}\right)+\left(\frac{1}{5 + 1}-\frac{1}{5+2}\right)+\left(\frac{1}{6 + 1}-\frac{1}{6+2}\right)\ &=\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)\ &=\frac{1}{5}-\frac{1}{8}\ &=\frac{8 - 5}{40}=\frac{3}{40} \end{align*} ]

Step2: Calculate (S_4)

[ \begin{align*} S_4&=\sum_{n = 4}^{7}\left(\frac{1}{n + 1}-\frac{1}{n+2}\right)\ &=\left(\frac{1}{4 + 1}-\frac{1}{4+2}\right)+\left(\frac{1}{5 + 1}-\frac{1}{5+2}\right)+\left(\frac{1}{6 + 1}-\frac{1}{6+2}\right)+\left(\frac{1}{7 + 1}-\frac{1}{7+2}\right)\ &=\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)\ &=\frac{1}{5}-\frac{1}{9}\ &=\frac{9-5}{45}=\frac{4}{45} \end{align*} ]

Step3: Calculate (S_5)

[ \begin{align*} S_5&=\sum_{n = 4}^{8}\left(\frac{1}{n + 1}-\frac{1}{n+2}\right)\ &=\left(\frac{1}{4 + 1}-\frac{1}{4+2}\right)+\left(\frac{1}{5 + 1}-\frac{1}{5+2}\right)+\left(\frac{1}{6 + 1}-\frac{1}{6+2}\right)+\left(\frac{1}{7 + 1}-\frac{1}{7+2}\right)+\left(\frac{1}{8 + 1}-\frac{1}{8+2}\right)\ &=\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{10}\right)\ &=\frac{1}{5}-\frac{1}{10}\ &=\frac{2 - 1}{10}=\frac{1}{10} \end{align*} ]

Step4: Calculate (S)

For the telescoping series (S=\sum_{n = 4}^{\infty}\left(\frac{1}{n + 1}-\frac{1}{n+2}\right)), as (n\to\infty), most terms cancel out. [ \begin{align*} S&=\lim_{k\to\infty}S_k\ &=\lim_{k\to\infty}\sum_{n = 4}^{k}\left(\frac{1}{n + 1}-\frac{1}{n+2}\right)\ &=\frac{1}{5} \end{align*} ]

Answer:

(S_3=\frac{3}{40},S_4=\frac{4}{45},S_5=\frac{1}{10},S=\frac{1}{5})