I have a progression that goes like this
S = $\frac{3}{19}+\frac{33}{19^2}+\frac{333}{19^3}+\frac{3333}{19^4}... \infty$
I tried solving this sum in this way
Multiply both sides by $3$
$3S$ = $\frac{9}{19}+\frac{99}{19^2}+\frac{999}{19^3}+\frac{9999}{19^4}...$
$3S$ = $\frac{10-1}{19}+\frac{10^2-1}{19^2}+\frac{10^3-1}{19^3}+\frac{10^4-1}{19^4}...$
Then separate those $10$s and $1$s
$3S$ = $(\frac{10}{19}+\frac{10^2}{19^2}+\frac{10^3}{19^3}+\frac{10^4}{19^4}...)$ - $(\frac{1}{19}+\frac{1}{19^2}+\frac{1}{19^3}+\frac{1}{19^4}...)$
Now, I know that for a geometric progression with it's common ratio being smaller than one
$S = a + ar + ar^2 + ar^3 ... \infty $ $\qquad( r<1)$
the sum of the series till infinity is
$S=\frac{a}{1-r}$
So, the above equation now becomes
$3S = \Biggl(\frac{\frac{10}{19}}{1-\frac{10}{19}}\Biggl) - \Biggl(\frac{1}{1-\frac{1}{19}}\Biggl)$
$3S = \Biggl(\frac{10}{9}\Biggl) - \Biggl(\frac{19}{18}\Biggl)$
$3S = \frac{1}{18}$
And so, $S = \frac{1}{54}$
Now, the second method that I learnt from my teacher was this one
S = $\frac{3}{19}+\frac{33}{19^2}+\frac{333}{19^3}+\frac{3333}{19^4}... \infty$
Divide both sides by 19
$\frac{S}{19}$ = $\frac{3}{19^2}+\frac{33}{19^3}+\frac{333}{19^4}+\frac{3333}{19^5}...$
Subtract this equation from the original one
$S-\frac{S}{19}$ = $\frac{3}{19}+\frac{30}{19^2}+\frac{300}{19^3}+\frac{3000}{19^4}...$
$\frac{18S}{19}$ = $\frac{3}{19}(\frac{1}{1}+\frac{10}{19}+\frac{100}{19^2}+\frac{1000}{19^3}...)$
Again, the equation has been reduced to a geometric progression with it's common ratio smaller than 1, and first term being 1
After multiplying both sides by 19 and then dividing both sides by 3, I get
$6S$ = $\frac{1}{1-\frac{10}{19}}$
$6S$ = $\frac{19}{9}$
And finally, this says $S = \frac{19}{54}$ which is different from the previously solved answer $\frac{1}{54}$
What I wish to know is why would different methods give different results, and in case the method is wrong, please correct me on the exact point.
EDIT: The mistake was my confusion in taking the first term for sum of geometric series. I took the first term of the series $\frac{1}{19}+\frac{1}{19^2}..$ as 1 instead of $\frac{1}{19}$. As a result, the 19 on the numerator remained in the equation to be subtracted from the other term $\frac{20}{18}$ and hence take away the 19 on the numerator.
In your approach, what is $\frac{1}{19} + \frac{1}{19^2} + \ldots$?
You claim it is $\frac{19}{18}$, but that's very much larger than $ \frac{1}{19}$, so you should be suspicious about that value.
That's the first error.
There are a few other arithmetic errors lying around. In particular, the conclusion should be $ S = \frac{ 19}{54}$, which both approaches yield (once corrected).