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Consider the given series,

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

Consider the series and rewrite it,

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

1. if L<1 series converges

2. if L>1 series diverges

if L= 1 the test is inconclusive

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

According to the convergence and divergence of geometric series.

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

2. for $\left|r\right|鈮�1$, the geometric series $\underset{k=0}{\overset{鈭�}{鈭�}}{r}^k$diverges.

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

Rewrite the given series as follows,

Hence the series (1) can be written as,

$\left(1+\frac{1}{10}+\frac{1}{100}+....\right)+\left(\frac{1}{5}+\frac{1}{50}+\frac{1}{500}+...\right)=\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{{10}^k}+\frac{1}{5}\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{{10}^k}$

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

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

hence, the series$\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{{10}^k}\mathrm{will}\mathrm{converge}\mathrm{to}\frac{1}{1-{\displaystyle \frac{1}{10}}}$.

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

$\begin{array}{rcl}\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{{10}^k}+\frac{1}{5}\underset{k=0}{\overset{鈭�}{鈭�}}\frac{1}{{10}^k}& =& \frac{10}{9}+\frac{1}{5}\left(\frac{10}{9}\right)\\ & =& \frac{10}{9}\left(1+\frac{1}{5}\right)\\ & =& \frac{10}{9}\left(\frac{6}{5}\right)\\ & =& \frac{4}{3}\end{array}$

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