Chapter 8.2, Problem 6E

To determine

The first 10 partial sums, the graph of the sequence of the terms and the sequence of the partial sums and determine if the series is convergent or divergent and hence find the sum of the series.

Answer to Problem 6E

The first 10 partial sums are: $4.90000$ , $8.33000$ , $10.73100$ , $12.41170$ , $13.58819$ , $14.41173$ , $14.98821$ , $15.39175$ , $15.67422$ and $15.87196$ .

The given series is convergent and the sum of the series is 16.33.

Explanation of Solution

Given: The given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{7^{n+1}}{{10}^n}}$ .

Concepts Used: Let ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ be any series. Then the $n^{th}$ partial sum of the series is given by-

  $s_n={\displaystyle \sum _{k=1}^n{a}_k=a_1+a_2+\mathrm{....}+a_n}$

If the sequence $\left\{s_n\right\}$ is convergent and ${\lim }_{n\to \infty }{s}_n=s$ as a real number, then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is convergent and the sum of the series is equal to s , i.e.-

  ${\displaystyle \sum _{n=1}^{\infty }{a}_n}=s$

If the sequence $\left\{s_n\right\}$ is divergent, then the series is also divergent.

Thus sum of a GP series with first term a and common ratio r is given by-

  $S_n=\frac{a(r^n-1)}{r-1}$ , when $\left|r\right|>1$

  $=\frac{a(1-r^n)}{1-r}$ , when $\left|r\right|<1$

Calculations: The first 10 terms of the given sequence are as follows-

  $a_1=4.90000$

  $a_2=3.43000$

  $a_3=2.40100$

  $a_4=1.68070$

  $a_5=1.17649$

  $a_6=0.82354$

  $a_7=0.57648$

  $a_8=0.40354$

  $a_9=0.28248$

  $a_{10}=0.19773$

The first 10 partial sums are as follows-

  $s_1={\displaystyle \sum _{n=1}^1\frac{7^{n+1}}{{10}^n}}=4.90000$

  $s_2={\displaystyle \sum _{n=1}^2\frac{7^{n+1}}{{10}^n}}=8.33000$

  $s_3={\displaystyle \sum _{n=1}^3\frac{7^{n+1}}{{10}^n}}=10.73100$

  $s_4={\displaystyle \sum _{n=1}^4\frac{7^{n+1}}{{10}^n}}=12.41170$

  $s_5={\displaystyle \sum _{n=1}^5\frac{7^{n+1}}{{10}^n}}=13.58819$

  $s_6={\displaystyle \sum _{n=1}^6\frac{7^{n+1}}{{10}^n}}=14.41173$

  $s_7={\displaystyle \sum _{n=1}^7\frac{7^{n+1}}{{10}^n}}=14.98821$

  $s_8={\displaystyle \sum _{n=1}^8\frac{7^{n+1}}{{10}^n}}=15.39175$

  $s_9={\displaystyle \sum _{n=1}^9\frac{7^{n+1}}{{10}^n}}=15.67422$

  $s_{10}={\displaystyle \sum _{n=1}^{10}\frac{7^{n+1}}{{10}^n}}=15.87196$

Hence, the graph of the terms of the sequence of the partial sums of the sequence is given by-

  

The $n^{th}$ partial sum of the series is given by-

  $s_n={\displaystyle \sum _{k=1}^n{a}_k}$

  $={\displaystyle \sum _{k=1}^n\frac{7^{k+1}}{{10}^k}}$

  $={\displaystyle \sum _{k=1}^n\frac{7\times 7^k}{{10}^k}}$

  $=7{{\displaystyle \sum _{k=1}^n\left(\frac{7}{10}\right)}}^k$ , which is a GP series with common ratio $\frac{7}{10}$ .

  $=7\times \frac{\frac{7}{10}\left[1-{\left(\frac{7}{10}\right)}^n\right]}{1-\left(\frac{7}{10}\right)}$

  $=\frac{49}{3}\times \left[1-{\left(\frac{7}{10}\right)}^n\right]$

Now,

  $\underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\left\{\frac{49}{3}\times \left[1-{\left(\frac{7}{10}\right)}^n\right]\right\}$

  $=\frac{49}{3}\times \underset{n\to \infty }{\lim }\left[1-{\left(\frac{7}{10}\right)}^n\right]$

  $=\frac{49}{3}$ (Since, $\underset{n\to \infty }{\lim }{\left(\frac{7}{10}\right)}^n=0$ )

  $=16.33$ (approx.)

Thus, the given series is convergent and it converges at $16.33$ , i.e.- the sum of the series is 16.33.