Question

For the series$\sum _{k=0}^{\mathrm{\infty }} \left(\frac{1}{k+3}-\frac{1}{k+4}\right)$that follow,

Part (a): Provide the first five terms in the sequence of partial sums $\left\{S_k\right\}$.

Part (b): Provide a closed formula for $S_k$.

Part (c): Find the sum of the series by evaluating$\underset{k\to \infty }{\lim }{S}_k$.

Short answer

Part (a): The first five terms of partial sums for the given series is$\left\{\frac{1}{12},\frac{2}{15},\frac{1}{6},\frac{4}{21},\frac{5}{25}\right\}$.

Part (b): The general term $S_k$in its sequence of partial sums is $\frac{1}{3}-\frac{1}{k+4}$.

Part (c): The sum of the series is$\frac{1}{3}$.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

$\sum _{k=0}^{\mathrm{\infty }} \left(\frac{1}{k+3}-\frac{1}{k+4}\right)$

Part (a) Step 2. Find the first two terms in the sequence.

The first term of the given series is obtained by substituting $k=0$,

$$\begin{gathered} =\frac{1}{0+3}-\frac{1}{0+4} \\ =\frac{1}{3}-\frac{1}{4} \\ =\frac{1}{12} \end{gathered}$$

First term is $\frac{1}{12}$.

The second term of the given series is obtained by substituting $k=1$,

$\begin{array}{r}=\frac{1}{1+3}-\frac{1}{1+4}\\ =\frac{1}{4}-\frac{1}{5}\\ =\frac{1}{20}\end{array}$

Second term is $\frac{1}{20}$.

Part (a) Step 3. Find the third, fourth terms in the sequence.

The third term of the given series is obtained by substituting $k=2$,

$\begin{array}{r}=\frac{1}{2+3}-\frac{1}{2+4}\\ =\frac{1}{5}-\frac{1}{6}\\ =\frac{1}{30}\end{array}$

Third term is $\frac{1}{30}$.

The fourth term of the given series is obtained by substituting $k=3$,

$\begin{array}{r}=\frac{1}{3+3}-\frac{1}{3+4}\\ =\frac{1}{6}-\frac{1}{7}\\ =\frac{1}{42}\end{array}$

Fourth term is$\frac{1}{42}$.

Part (a) Step 4. Find the fifth terms in the sequence.

The fifth term of the given series is obtained by substituting $k=4$,

$\begin{array}{r}=\frac{1}{4+3}-\frac{1}{4+4}\\ =\frac{1}{7}-\frac{1}{8}\\ =\frac{1}{56}\end{array}$

Fifth term is $\frac{1}{56}$.

The first and second terms in the sequence of partial sum is given below,

$$\begin{gathered} S_1=\frac{1}{3}-\frac{1}{4} \\ =\frac{1}{12} \\ {S}_2=S_1+a_2 \\ =\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5} \\ =\frac{2}{15} \end{gathered}$$

Part (a) Step 5. Find the partial sums.

The third, fourth and fifth terms in the sequence of partial sum is given below,

$$\begin{gathered} S_3=S_2+a_3 \\ =\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6} \\ {S}_4=S_3+a_4 \\ =\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7} \\ =\frac{4}{21} \\ {S}_5=S_4+a_5 \\ =\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8} \\ =\frac{5}{24} \end{gathered}$$

Part (b) Step 1. Write a close formula for Sk.

The kth term in the sequence of the partial sums is given below,

$S_k=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)+...+\left(\frac{1}{k+3}-\frac{1}{k+4}\right)$

In each two consecutive pairs, the second term of a pair cancels with the first term of the subsequent pair.

Thus, the series is telescopic.

The general term in its sequence of partial sums is $S_k=\frac{1}{3}-\frac{1}{k+4}$.

Part (c) Step 1. Find the sum of the series.

The $S_k$in its sequence of partial sums is $S_k=\frac{1}{3}-\frac{1}{k+4}$.

The value of $\underset{k\to \infty }{\lim }{S}_k$is given below,

$$\begin{gathered} \underset{k\to \infty }{\lim }{S}_k=\underset{k\to \infty }{\lim }\left(\frac{1}{3}-\frac{1}{k+4}\right) \\ =\frac{1}{3}-0 \\ =\frac{1}{3} \end{gathered}$$