Let we are given a series as $S=\dfrac{4}{19}+\dfrac{44}{{{19}^{2}}}+\dfrac{444}{{{19}^{3}}}+\dfrac{4444}{{{19}^{4}}}+................\text{up to }\infty $, then ‘S’ is equal to ?

Hint: The given series is not one of the standard series, that is, it is not an AP or GP series. But, it can be converted into one by some manipulation. In the denominator of our series, we can see the ratio is constant and equal to 19. Thus, our manipulations will be in such a way that we convert this series into an infinite geometric progression, whose sum can then be easily calculated.

Complete step-by-step solution:
In our problem, we have been given the progression whose sum is to be calculated as:
$\Rightarrow S=\dfrac{4}{19}+\dfrac{44}{{{19}^{2}}}+\dfrac{444}{{{19}^{3}}}+\dfrac{4444}{{{19}^{4}}}+................\text{up to }\infty $
Let us say the above equation is equation number (1). So, we have:
$\Rightarrow S=\dfrac{4}{19}+\dfrac{44}{{{19}^{2}}}+\dfrac{444}{{{19}^{3}}}+\dfrac{4444}{{{19}^{4}}}+................\text{up to }\infty $ .......... (1)
Now, dividing both sides of equation number (1) by 19, we get:
$\Rightarrow \dfrac{S}{19}=\dfrac{4}{{{19}^{2}}}+\dfrac{44}{{{19}^{3}}}+\dfrac{444}{{{19}^{4}}}+\dfrac{4444}{{{19}^{5}}}+................\text{up to }\infty $
Let us say the above equation is equation number (2). So, we have:
$\Rightarrow \dfrac{S}{19}=\dfrac{4}{{{19}^{2}}}+\dfrac{44}{{{19}^{3}}}+\dfrac{444}{{{19}^{4}}}+\dfrac{4444}{{{19}^{5}}}+................\text{up to }\infty $ ............. (2)

Now, on subtracting equation number (2) by (1), we get the new progression as:
$\begin{align}
  & \Rightarrow S-\dfrac{S}{19}=\left[ \dfrac{4}{19}+\dfrac{44}{{{19}^{2}}}+\dfrac{444}{{{19}^{3}}}+\dfrac{4444}{{{19}^{4}}}+.......\text{up to }\infty \right]-\left[ \dfrac{4}{{{19}^{2}}}+\dfrac{44}{{{19}^{3}}}+\dfrac{444}{{{19}^{4}}}+\dfrac{4444}{{{19}^{5}}}+.......\text{up to }\infty \right] \\
 & \Rightarrow \dfrac{18S}{19}=\dfrac{4}{19}+\left( \dfrac{44}{{{19}^{2}}}-\dfrac{4}{{{19}^{2}}} \right)+\left( \dfrac{444}{{{19}^{3}}}-\dfrac{44}{{{19}^{3}}} \right)+\left( \dfrac{4444}{{{19}^{4}}}-\dfrac{444}{{{19}^{4}}} \right)+............\text{up to }\infty \\
\end{align}$
$\Rightarrow \dfrac{18S}{19}=\dfrac{4}{19}+\dfrac{40}{{{19}^{2}}}+\dfrac{400}{{{19}^{3}}}+\dfrac{4000}{{{19}^{4}}}+...........\text{ up to }\infty $
Here, we can clearly see that our final progression is a geometric progression with its constant ratio as $\dfrac{10}{19}$. This ratio is less than 1, so we can apply the summation formula for the series sum of an infinite GP. This formula is given as follows:
$\Rightarrow {{S}_{\infty GP}}=\dfrac{a}{1-r}$
Where,
‘a’ is the first term of the series. And,
‘r’ is the common ratio of the infinite GP series.
Using the above formula in our progression, we get:
$\begin{align}
  & \Rightarrow \dfrac{18S}{19}=\dfrac{\dfrac{4}{19}}{1-\dfrac{10}{19}} \\
 & \Rightarrow \dfrac{18S}{19}=\dfrac{\dfrac{4}{19}}{\dfrac{9}{19}} \\
 & \Rightarrow \dfrac{18S}{19}=\dfrac{4}{9} \\
 & \Rightarrow S=\dfrac{19\times 4}{18\times 9} \\
 & \therefore S=\dfrac{38}{81} \\
\end{align}$
Hence, the value of “S” comes out to be $\dfrac{38}{81}$.

Note: In problems, where the series is not in a standard form, we should always try different manipulations to convert the series into a standard series. These manipulations can be finding the general term of the series or like the one that we used in our problem, depending on the question. Also, one should always re-check their solution for any possible error that may have concurred.