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.

Short answer

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

Step-by-step solution

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.

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.

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.

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.

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 $\frac{4}{3}$.