Is It A Converging Series?

$\begin{aligned} S & = 0.1+0.04+0.009+0.0016+... = \sum_{n=1}^\infty \frac {n^2}{10^n} \end{aligned}$

Note that $\displaystyle \sum_{n=1}^\infty \frac {n^2}{10^n} = \sum_{n=\color{#D61F06}{0}}^\infty \frac {n^2}{10^n}$ . So we have:

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {n^2}{10^n} \\ & = \sum_{n=\color{#D61F06}{0}}^\infty \frac {({\color{#D61F06}n+1})^2}{10^{\color{#D61F06}{n+1}}} \\ & = \sum_{n=\color{#D61F06}{0}}^\infty \frac {n^2+2n+1}{10^{n+1}} \\ & = \frac 1{10} \sum_{n=0}^\infty \frac {n^2}{10^n} + \frac 15 {\color{#3D99F6}\sum_{n=0}^\infty \frac {n}{10^n}} + \frac 1{10} \sum_{n=0}^\infty \frac 1{10^n} & \small {\color{#3D99F6} \text{see Note}} \\ & = \frac 1{10} S + \frac 15 {\color{#3D99F6}\left(\frac {10}{81}\right)} + \frac 1{10} \left(\frac 1{1-\frac 1{10}} \right) \\ \implies \frac 9{10} S & = \frac 2{81} + \frac 19 = \frac {11}{81} \\ \implies S & = \frac {110}{729} \end{aligned}$

$\implies a+b = 110+729 = \boxed{839}$

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Note:

Using the similar method:

$\begin{aligned} S_1 & = \sum_{n=0}^\infty \frac {n}{10^n} \\ & = \sum_{n=\color{#D61F06}{1}}^\infty \frac {n}{10^n} \\ & = \sum_{n=\color{#3D99F6}{0}}^\infty \frac {\color{#3D99F6}{n+1}}{10^{\color{#3D99F6}{n+1}}} \\ & = \frac 1{10} \sum_{n=0}^\infty \frac {n}{10^n} + \frac 1{10} \sum_{n=0}^\infty \frac 1{10^n} \\ & = \frac 1{10} S_1 + \frac 1{10} \left(\frac 1{1-\frac 1{10}} \right) \\ \implies \frac 9{10} S & = \frac 19 \\ \implies S & = \frac {10}{81} \end{aligned}$

$S=0.1+2^2(0.1)^2+3^2(0.1)^3+... \infty$

Let $x=0.1$ , then:

$S=x+2^2x^2+3^2x^3+... \infty$

$S=x + (2x^2 + 2x^2) + (3x^3 + 3x^3 + 3x^3) + ... \infty$

$S=(x+2x^2+3x^3+...\infty)+(2x^2+3x^3+4x^4+...\infty)+(3x^3+4x^4+5x^5+...\infty)+...\infty$

$S=(2x-x+3x^2-x^2+4x^3-x^3+...\infty)+(3x^2-x^2+4x^3-x^3+...\infty)+(4x^3-x^3+5x^4-x^4+6x^5-x^5+...\infty)+...\infty$

$=\left[\frac{d}{dx}\left(x^2+x^3+...\infty \right)\right]-\left(x+x^2+x^3+...\infty \right) + \left[\frac{d}{dx}\left(x^3+x^4+...\infty \right)\right]-\left(x^2+x^3+x^4+...\infty \right) + \left[\frac{d}{dx}\left(x^4+x^5+...\infty \right)\right]-\left(x^3+x^4+x^5+...\infty\right)+...\infty$

Since $|x|<1$ , we can use the formula of infinite geometric progression sum:

$=\left[\frac{d}{dx}\left(\frac{x^2+x^3+x^4+...}{1-x}\right)\right]-\frac{x+x^2+x^3+...}{1-x}$

$=\left[\frac{d}{dx}\left(\frac{x^2}{\left(1-x\right)^2}\right)\right]-\frac{x}{\left(1-x\right)^2}$

$=\frac{x(1+x)}{(1-x)^3} = \frac{0.1(1+0.1)}{(1-0.1)^3}=\boxed{\frac{110}{729}}$