Q. 65 - Calculus

Question

Demonstrate that each of the series  is telescoping. Give the general term Sn in each series' list of partial sums, and if the series converges, get the total of the series.

Step-by-Step Solution

Answer

 The series $\sum _{k=1}^{\infty }\ln \frac{k}{k+1}$ is divergent

Step 1: Given information

The series is $\sum _{k=1}^{\infty }\ln \frac{k}{k+1}.$

Step 2: Calculation

The goal is to demonstrate the series' telescopic nature, supply the general term S n in its list of partial sums, and determine whether the series is convergent by determining its sum.

If each term of the series can be expressed as the sum of two or more summands and there is cancellation between terms in the sequence of partial sums, the series is telescopic.

The series $\sum _{k=1}^{\infty }\ln \frac{k}{k+1}$ can be expressed as

$\sum _{k=1}^{\infty }\ln \frac{k}{k+1}=\sum _{k=1}^{\infty }[\ln k-\ln (k+1\left)\right](Because\left.\ln \frac{a}{b}=\ln a-\ln b\right)$

In the series of partial sums, the nth term is

$S_n=\ln \left(1\right)-\ln \left(2\right)+(\ln 2-\ln 3)+\dots (\ln n-\ln (n+1\left)\right)$

The second term of a pair cancels with the first term of the following pair in each pair of two successive pairs.

The series is therefore telescopic.

When it comes to its series of partial sums, the general term S n is

$S_n=\ln \left(1\right)-\ln \left(2\right)+(\ln 2-\ln 3)+\dots (\ln n-\ln (n+1\left)\right)$

$$\begin{gathered} =\ln \left(1\right)-\ln \left(2\right)+(\ln 2-\ln 3)+(\ln 3-\ln 4)\dots +(\ln n-\ln (n+1\left)\right) \\ =\ln \left(1\right)-\ln (n+1) \\ =-\ln (n+1)(\text{ Because }\ln 1=0) \\ =\ln \left(\frac{1}{n+1}\right) \end{gathered}$$

Consequently, the generic term S n in its series of partial sums is

$\ln \left(\frac{1}{n+1}\right).$

The upper bound for S n as n $\to \infty$ is

$$\begin{gathered} \underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\left[\ln \left(\frac{1}{n+1}\right)\right] \\ =\infty \end{gathered}$$