How do you find the sum of the infinite series $ \sum\limits_{i=1}^{\infty }{6{{\left( \dfrac{1}{10} \right)}^{i}}} $ ?

Hint: We first get the terms using the general terms of the series. Then we find the first term $ {{t}_{1}} $ and common ratio $ r $ . We use the formula of sum of infinite G.P. to find the final solution for $ \sum\limits_{i=1}^{\infty }{6{{\left( \dfrac{1}{10} \right)}^{i}}} $.

Complete step by step solution:
The given series $ \sum\limits_{i=1}^{\infty }{6{{\left( \dfrac{1}{10} \right)}^{i}}} $ is an infinite G.P. series. We can write it as $ \sum\limits_{i=1}^{\infty }{6{{\left( \dfrac{1}{10} \right)}^{i}}}=6\sum\limits_{i=1}^{\infty }{{{\left( \dfrac{1}{10} \right)}^{i}}} $ .
We express the geometric sequence in its general form.
We express the terms as $ {{t}_{n}} $ , the $ {{n}^{th}} $ term of the series $ \sum\limits_{i=1}^{\infty }{{{\left( \dfrac{1}{10} \right)}^{i}}} $ .
Now we place consecutive natural numbers for $ i $ as $ 1,2,3,4,... $ to get the sequence as
\[{{\left( \dfrac{1}{10} \right)}^{1}},{{\left( \dfrac{1}{10} \right)}^{2}},{{\left( \dfrac{1}{10} \right)}^{3}},{{\left( \dfrac{1}{10} \right)}^{4}},......\]. The first term be $ {{t}_{1}} $ and the common ratio be $ r $ where $ r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}} $ . Here $ {{t}_{1}}=\left( \dfrac{1}{10} \right) $ and $ r=\left( \dfrac{1}{10} \right)<1 $ .
We can express the general term $ {{t}_{n}} $ based on the first term and the common ratio.
The formula being $ {{t}_{n}}={{t}_{1}}{{r}^{n-1}} $ .
If we assume the sum of the infinite G.P as $ {{S}_{\infty }} $ then for the value of $ \left| r \right|<1 $ , the sum of the infinite terms of the G.P. is $ {{S}_{\infty }}=\dfrac{{{t}_{1}}}{1-r} $ .
Putting the values, we get \[{{S}_{\infty }}=\sum\limits_{i=1}^{\infty }{{{\left( \dfrac{1}{10} \right)}^{i}}}=\dfrac{\dfrac{1}{10}}{1-\dfrac{1}{10}}=\dfrac{1}{10}\times \dfrac{10}{9}=\dfrac{1}{9}\].
The final solution is $ \sum\limits_{i=1}^{\infty }{6{{\left( \dfrac{1}{10} \right)}^{i}}}=6\sum\limits_{i=1}^{\infty }{{{\left( \dfrac{1}{10} \right)}^{i}}}=\dfrac{6}{9}=\dfrac{2}{3} $ .
So, the correct answer is “$\dfrac{2}{3} $”.

Note: The sequence is a decreasing sequence where the common ratio is a value of less than 1. The common difference will never be calculated according to the difference of greater number from the lesser number. The ratio formula should always be according $ r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}} $ .