Chapter 12.1, Problem 40E

To determine

Calculate the first four partial sums and also find the $n^{th}$ partial sum of the sequence $a_n$ .

Answer to Problem 40E

The first four partial sums is,

  $S_1=\log \left(\frac{1}{2}\right),S_2=\log \left(\frac{1}{3}\right),S_3=\log \left(\frac{1}{4}\right),S_4=\log \left(\frac{1}{5}\right)$

And, the $n^{th}$ partial sum of the sequence is,

  $S_n=\log \left(\frac{1}{n+1}\right)$

Explanation of Solution

Given information:

It is given in the question that, $a_n=\log \left(\frac{n}{n+1}\right)$ .

Formula used:

For finding the partial sums use the formula,

  $S_n=a_1+a_2+a_3+\mathrm{........}+a_n$ .

Calculation: Since, $a_n=\log \left(\frac{n}{n+1}\right)$ so now by putting the value of $n=1,2,3,4$ the first four term of $A.P$ are arrived

That is,

  $a_1=\log \left(\frac{1}{2}\right)=\log (1)-\log (2)$ (by using the property of log)

  $a_2=\log \left(\frac{2}{3}\right)=\log (2)-\log (3)$

  $a_3=\log \left(\frac{3}{4}\right)=\log (3)-\log (4)$

  $a_4=\log \left(\frac{4}{5}\right)=\log (4)-\log (5)$

For finding the $n^{th}$ partial sum of the sequence use the formula $S_n=a_1+a_2+a_3+\mathrm{........}+a_n$

  $\begin{array}{l}{S}_1=\log (1)-\log (2)\\ S_2=\log (1)-\log (3)\\ S_3=\log (1)-\log (4)\\ S_4=\log (1)-\log (5)\end{array}$

Finally, by using the property of log,

  $S_1=\log \left(\frac{1}{2}\right),S_2=\log \left(\frac{1}{3}\right),S_3=\log \left(\frac{1}{4}\right),S_4=\log \left(\frac{1}{5}\right)$

By observing the above terms,

  $S_n=\log \left(\frac{1}{n+1}\right)$