Infinite Geometric Series Sum The sum to infinity of the series $$1+\frac2{11}+\frac3{121}+\frac4{1331}+\cdots$$
I tried finding the common ratio but I was not able to find it. I tried putting it in closed form but even that was not enough. Please help.
 A: Let $f(t)= \sum\limits_{n=0}^{\infty} t^{n+1}$. Then $f'(t)=\sum\limits_{n=0}^{\infty}(n+1) t^{n}$. The given sum is nothing but $f'(\frac  1 {11})$. Use the fact that $f(t)=\frac t {1-t}$to compute $f'(\frac  1 {11})$.
A: An arithmetico-geometric sequence is composed of both an arithmetic progression and a geometric progression. You can notice that in your given sequence: the numerator has a constant difference whereas the denominator has a constant ratio. Given the initial value $a$ and the constant difference of $d$ in the numerator, and initial value $b$ and the constant ratio $r$ in the denominator, the sum of the infinite series is given by
$$\frac{ab}{1 - r} + \frac{dbr}{\left(1 - r\right)^2}.$$
In your case, $a = b = d = 1$ and $r = \dfrac1{11}$. Therefore, the sum of the infinite series is $$\frac{1\cdot1}{1 - \frac1{11}} + \frac{1\cdot1\cdot\frac1{11}}{\left(1 - \frac1{11}\right)^2} = \frac{11}{10}+\frac{11}{100} = \frac{121}{100}$$ 
A: Denote:
$$S=1+\frac2{11}+\frac3{121}+\frac4{1331}+\cdots \Rightarrow \\
11S=11+\color{red}1+\color{blue}1+\frac{\color{red}1+\color{blue}2}{11}+\frac{\color{red}1+\color{blue}3}{121}+\cdots=11+\left(\color{red}{1+\frac1{11}+\frac1{121}+\cdots}\right)+\color{blue}S \Rightarrow \\
10S=11+\color{red}{\frac{1}{1-\frac1{11}}}=11+\frac{11}{10}=\frac{121}{10} \Rightarrow S=\frac{121}{100}.$$
