91Ó°ÊÓ

Consider the given question,

$\underset{k=2}{\overset{\mathrm{∞}}{∑}} \ln \left(\frac{k}{k+1}\right)$

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

$$\begin{gathered} =\ln k−\ln (k+1) \\ =\ln 2−\ln 3 \end{gathered}$$

First term is $\ln 2−\ln 3$.

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

$$\begin{gathered} =\ln k−\ln (k+1) \\ =\ln 3−\ln 4 \end{gathered}$$

Second term is $\ln 3−\ln 4$.

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

$$\begin{gathered} =\ln k−\ln (k+1) \\ =\ln 4−\ln 5 \end{gathered}$$

The fourth term of the given series is obtained by substituting $k=5$,

$$\begin{gathered} =\ln k−\ln (k+1) \\ =\ln 5−\ln 6 \end{gathered}$$

Fourth term is $\ln 5−\ln 6$.

The fifth term of the given series is obtained by substituting $k=6$,

$$\begin{gathered} =\ln k−\ln (k+1) \\ =\ln 6−\ln 7 \end{gathered}$$

Fifth term is $\ln 6−\ln 7$.

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

$$\begin{gathered} S_1=\ln 2−\ln 3 \\ {S}_2=S_1+a_2 \\ =\ln 2−\ln 3+\ln 3−\ln 4 \\ =\ln 2−\ln 4 \end{gathered}$$

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

$$\begin{gathered} S_3=S_2+a_3 \\ =\ln 2−\ln 4+\ln 4−\ln 5 \\ =\ln 2−\ln 5+\ln 5−\ln 6 \\ =\ln 2−\ln 6 \\ {S}_5=S_4+a_5 \\ =\ln 2−\ln 6+\ln 6−\ln 7 \end{gathered}$$

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

$S_k=\left(\ln 2-\ln 3\right)+\left(\ln 3-\ln 4\right)+...+\ln \left(\frac{k}{k+1}\right)$

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$S_k=\log \left(\mathrm{Γ}\right(n+1\left)\right)−\log \left(\mathrm{Γ}\right(n+2\left)\right)+\log \left(2\right)$.

The $S_k$in its sequence of partial sums is $S_k=\log \left(\mathrm{Γ}\right(n+1\left)\right)−\log \left(\mathrm{Γ}\right(n+2\left)\right)+\log \left(2\right)$.

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

$\underset{k→∞}{\lim }{S}_k=\underset{k→∞}{\lim }\log \left(\mathrm{Γ}\right(n+1\left)\right)−\log \left(\mathrm{Γ}\right(n+2\left)\right)+\log \left(2\right)$