Question

For the series $\sum _{k=2}^{\mathrm{\infty }} \ln \left(\frac{k}{k+1}\right)$that follow,

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

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 .

Part (b): The general term $S_k$in its sequence of partial sums is .

Part (c): The sum of the series is $\underset{k\to \infty }{\lim }\log \left(\mathrm{\Gamma }\right(n+1\left)\right)-\log \left(\mathrm{\Gamma }\right(n+2\left)\right)+\log \left(2\right)$.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

$\sum _{k=2}^{\mathrm{\infty }} \ln \left(\frac{k}{k+1}\right)$

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

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

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

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 $\ln 3-\ln 4$.

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

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 $\ln 5-\ln 6$.

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

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,

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

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

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

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

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

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

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

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

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

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

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

$\underset{k\to \infty }{\lim }{S}_k=\underset{k\to \infty }{\lim }\log \left(\mathrm{\Gamma }\right(n+1\left)\right)-\log \left(\mathrm{\Gamma }\right(n+2\left)\right)+\log \left(2\right)$