Q.19 - Calculus

Question

explain why the ratio test cannot be used on the series $1+\frac{1}{5}+\frac{1}{10}+\frac{1}{50}+\frac{1}{100}+\frac{1}{500}+....$ then show that the series converges and find its sum.

Step-by-Step Solution

Answer

The required sum to converges the series is $\frac{4}{3}$

Step 1: Given information

Consider the given series,

$1+\frac{1}{5}+\frac{1}{10}+\frac{1}{50}+\frac{1}{100}+\frac{1}{500}+.....$

To explain why the ratio test cannot be used on the given series. to show that the series converges.

Step 2 : Calculation

Consider the series and rewrite it,

According to the ratio test $\sum _{k=1}^{\infty }{a}_k$ be the series with all terms positive and $L=\underset{k\to \infty }{\lim }\left(\frac{a_{k+1}}{a_k}\right)$ then,

1. if L<1 series converges 

2. if L>1 series diverges

if L = 1 the test is inconclusive

Since $L=\underset{k\to \infty }{\lim }\left(\frac{a_{k+1}}{a_k}\right)$ does not exists this implies that the ratio test cannot be used.

Step 3: Further Calculation.

According to the convergence and divergence of geometric series.

 

1. for $\left|r\right|<1$, the geometric series $\sum _{k=0}^{\infty }{r}^k$converges to the sum $\frac{1}{1-r}$

2. for $\left|r\right|\ge 1$, the geometric series $\sum _{k=0}^{\infty }{r}^k$ diverges.

Step 4 : Simplification

Now the series,$\left(1+\frac{1}{10}+\frac{1}{100}+....\right)+\left(\frac{1}{5}+\frac{1}{50}+\frac{1}{500}+...\right)\to \left(1\right)$.

Rewrite the given series as follows, 

$\begin{array}{rcl}\left(1+\frac{1}{10}+\frac{1}{100}+....\right)& =& \sum _{k=0}^{\infty }\frac{1}{{10}^k}\\ \frac{1}{5}\left(1+\frac{1}{10}+\frac{1}{100}+....\right)& =& \frac{1}{5}\sum _{k=0}^{\infty }\frac{1}{{10}^k}\end{array}$

Hence the series (1) can be written as,

$\left(1+\frac{1}{10}+\frac{1}{100}+....\right)+\left(\frac{1}{5}+\frac{1}{50}+\frac{1}{500}+...\right)=\sum _{k=0}^{\infty }\frac{1}{{10}^k}+\frac{1}{5}\sum _{k=0}^{\infty }\frac{1}{{10}^k}$

Hence the series is convergent as its sum of two convergent geometric series.

Step 5 : Find the sum of the obtained series.

The sum of the convergent geometric series is $\frac{1}{1-r}$.

hence, the series$\sum _{k=0}^{\infty }\frac{1}{{10}^k} \mathrm{will} \mathrm{converge} \mathrm{to} \frac{1}{1-{\displaystyle \frac{1}{10}}}$.

Hence the sum of the series can be calculated as follows,

 

$\begin{array}{rcl}\sum _{k=0}^{\infty }\frac{1}{{10}^k}+\frac{1}{5}\sum _{k=0}^{\infty }\frac{1}{{10}^k}& =& \frac{10}{9}+\frac{1}{5}\left(\frac{10}{9}\right)\\ & =& \frac{10}{9}\left(1+\frac{1}{5}\right)\\ & =& \frac{10}{9}\left(\frac{6}{5}\right)\\ & =& \frac{4}{3}\end{array}$

Hence the sum of the series is $\frac{4}{3}$.