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First note that $\dfrac{S}{10} = \displaystyle\sum_{n=2}^{\infty} \dfrac{(n - 1) \times n^{2}}{10^{n}} \lt \sum_{n=2}^{\infty} \dfrac{n^{3}}{10^{n}}$ , a series which converges by the ratio test , implying that $S$ must converge as well. We are thus free to rearrange the terms of $S$ without changing its value.

We can then state that

$\dfrac{S}{10} = \displaystyle\sum_{n=2}^{\infty} \dfrac{n^{3}}{10^{n}} - \sum_{n=2}^{\infty} \dfrac{n^{2}}{10^{n}} = \left(\sum_{n=1}^{\infty} \dfrac{n^{3}}{10^{n}} - \dfrac{1}{10}\right) - \left( \sum_{n=1}^{\infty} \dfrac{n^{2}}{10^{n}} - \dfrac{1}{10}\right) = \sum_{n=1}^{\infty} \dfrac{n^{3}}{10^{n}} - \sum_{n=1}^{\infty} \dfrac{n^{2}}{10^{n}}.$

Next, we will make use of the fact that $\displaystyle\sum_{n=0}^{\infty} x^{n} = \dfrac{1}{1 - x}$ for $|x| \lt 1.$

Differentiating this equation, (the LHS term-by-term), we find that

$\displaystyle\sum_{n=1}^{\infty} nx^{n-1} = \dfrac{1}{(1 - x)^{2}} \Longrightarrow \sum_{n=1}^{\infty} nx^{n} = \dfrac{x}{(1 - x)^{2}}.$

Differentiating this last equation, (the LHS again term-by-term), we find that

$\displaystyle\sum_{n=1}^{\infty} n^{2} x^{n-1} = \dfrac{1 + x}{(1 - x)^{3}} \Longrightarrow \sum_{n=1}^{\infty} n^{2}x^{n} = \dfrac{x^{2} + x}{(1 - x)^{3}},$ (equation I).

Differentiating one last time, we find that

$\displaystyle\sum_{n=1}^{\infty} n^{3}x^{n-1} = \dfrac{x^{2} + 4x + 1}{(1 - x)^{4}} \Longrightarrow \sum_{n=1}^{\infty} n^{3}x^{n} = \dfrac{x^{3} + 4x^{2} + x}{(1 - x)^{4}}$ , (equation II).

So by plugging in $x = \dfrac{1}{10}$ into equations I and II, we see that

$\dfrac{S}{10} = \dfrac{\frac{1}{1000} + \frac{4}{100} + \frac{1}{10}}{(1 - \frac{1}{10})^{4}} - \dfrac{\frac{1}{100} + \frac{1}{10}}{(1 - \frac{1}{10})^{3}} = \dfrac{10^{4}}{9^{4}}*\dfrac{141}{10^{3}} - \dfrac{10^{3}}{9^{3}}*\dfrac{11}{100} = \dfrac{1410}{9^{4}} - \dfrac{110}{9^{3}} = \dfrac{420}{6561} = \dfrac{140}{2187}.$

Thus $S = \dfrac{1400}{2187}$ , and so $A + B = 1400 + 2187 = \boxed{3587}.$

A purely algebraic method: $S = \displaystyle \sum_{n=1}^\infty \frac {n(n+1)^2}{10^n}$ Note that $\frac {(n+1)(n+2)^2}{10^{n+1}} - \frac {n(n+1)^2}{10^{n+1}} = \frac {(n+1)(3n+4)}{10^{n+1}}$ We subtract $\frac {1}{10}S$ from $S$ to get $S- \frac {1}{10}S = \frac {2}{5} + \displaystyle \sum_{n=1}^\infty \frac {(n+1)(3n+4)}{10^{n+1}}$ Let $x = \displaystyle \sum_{n=1}^\infty \frac {(n+1)(3n+4)}{10^{n+1}}$ . Using the same method and infinite G.P. formula, $x- \frac {1}{10}x = \frac {7}{50} + \displaystyle \sum_{n=1}^\infty \frac {6n+10}{10^{n+2}} = \frac {7}{50} + \displaystyle \sum_{n=1}^\infty \frac {10}{10^{n+2}} + \displaystyle \sum_{n=1}^\infty \frac {6n}{10^{n+2}} \Rightarrow \frac {9}{10}x = \frac {7}{50} + \frac {1}{90} + \displaystyle \sum_{n=1}^\infty \frac {6n}{10^{n+2}}$ Let $y = \displaystyle \sum_{n=1}^\infty \frac {6n}{10^{n+2}}$ Using the method a third and final time together with the G.P. formula, $y - \frac {1}{10}y = \frac {6}{10^3} + \frac {6}{10^4} + \frac {6}{10^5} + \ldots = \frac {6}{900} = \frac {1}{150}$ Now we can solve for S from the 3 equations $\frac {9}{10} S = \frac {2}{5} + x$ , $\frac {9}{10} x = \frac {7}{50} + \frac {1}{90} + y$ and $\frac {9}{10}y = \frac {1}{150}$ to get $S = \frac {1400}{2187}$ . The required answer is then $1400+2187 = \boxed {3587}$

$S=\sum_{n=1}^{\infty} \frac{n(n+1)^2}{10^n}=\sum_{n=1}^{\infty} \frac{n^3+2n^2+n}{10^n}=\sum_{n=1}^{\infty} \frac{n^3}{10^n}+2\sum_{n=1}^{\infty} \frac{n^2}{10^n}+\sum_{n=1}^{\infty} \frac{n}{10^n}$

Using the known expressions of the polylogarithm we find: $S=Li_{-3}\left(\frac{1}{10}\right)+2Li_{-2}\left(\frac{1}{10}\right)+Li_{-1}\left(\frac{1}{10}\right)=\frac{1400}{2187}$