91Ó°ÊÓ

Consider the given question,

$\underset{k=0}{\overset{\mathrm{∞}}{∑}} \left(\frac{1}{k+3}−\frac{1}{k+4}\right)$

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 $\frac{1}{20}$.

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$\frac{1}{42}$.

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,

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

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

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

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 $S_k=\frac{1}{3}-\frac{1}{k+4}$.

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

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

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