Calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using pen and paper How can you calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using nothing more than a pen and pencil? Simply typing this in any symbolic calculator will give us $1/81$. I could also possibly find this formula if I was actually looking at given numbers but I have never tried working backwards. By backwards, I mean to be given the summation formula and determine the convergent limit for it (if it exists). 
So supposing that we did not know the limit for this summation formula was $1/81$ and that we do not have any software for assistance, how do we calculate this summation formula without having to take this summation to infinity?
 A: Hint: 
$$f(x) = \frac{1}{1 - x} = \sum\limits_{k = 0}^{\infty} x^k$$
and so differentiating term-by-term,
$$\frac{1}{(1 - x)^2} = \sum\limits_{k = 0}^{\infty} k x^{k - 1} = \sum\limits_{n = 1}^{\infty} (n - 1) x^n$$
A: T. Bongers has provided a general method, but I hope the following can help you understand it better. Write 
$$\sum_{n=1}^{\infty}\frac{(n-1)}{10^n}=\frac{(1-1)}{10}+\frac{(2-1)}{10^2}+\frac{(3-1)}{10^3}+\cdots\tag{1} $$ Therefore,
$$10\cdot\sum_{n=1}^{\infty}\frac{(n-1)}{10^n}=(1-1)+\frac{(2-1)}{10}+\frac{(3-1)}{10^2}+\cdots\tag{2}$$
Now subtract $(1)$ from $(2)$. As a result, we get
$$\begin{align*}
9\cdot\sum_{n=1}^{\infty}\frac{(n-1)}{10^n}&=\frac{(2-1)-(1-1)}{10}+\frac{(3-1)-(2-1)}{10^2}+\frac{(4-1)-(3-1)}{10^3}+\cdots\\
&=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\cdots\\
&=\frac{1}{1-1/10}-1\\&=\frac{1}{9}\end{align*}$$ Hence, $$\sum_{n=1}^{\infty}\frac{(n-1)}{10^n}=\frac{1}{9}\cdot\frac{1}{9}=\frac{1}{81}$$
A: We know that
$$
\begin{align}
A&=\hphantom{1+}\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots\\
10A&=1+\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots\\[4pt]
9A&=1
\end{align}
$$
So $\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots=A=\frac19$.
Likewise,
$$
\begin{align}
\color{#C00000}{\frac1{10}}A&=\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\frac1{10^5}+\dots\\
\color{#C00000}{\frac1{10^2}}A&=\hphantom{\frac1{10^2}+}\frac1{10^3}+\frac1{10^4}+\frac1{10^5}+\dots\\
\color{#C00000}{\frac1{10^3}}A&=\hphantom{\frac1{10^2}+\frac1{10^3}+}\frac1{10^4}+\frac1{10^5}+\dots\\
\color{#C00000}{\frac1{10^4}}A&=\hphantom{\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+}\frac1{10^5}+\dots\\
&\vdots\quad\text{summing the equations above}\\
\color{#C00000}{A}\hphantom{A}A&=\frac1{10^2}+\frac2{10^3}+\frac3{10^4}+\frac4{10^5}+\dots
\end{align}
$$
So $\frac1{10^2}+\frac2{10^3}+\frac3{10^4}+\frac4{10^5}+\dots=A^2=\frac1{81}$.
A: $$\sum\limits_{n=1}^\infty \frac{n-1}{10^n} = \sum\limits_{n=1}^\infty \sum\limits_{i=1}^{n-1} \frac{1}{10^n}= \sum\limits_{i=1}^\infty \sum\limits_{n=i+1}^\infty \frac{1}{10^{n}} = \sum\limits_{i=1}^\infty\frac{10^{-(i+1)}}{1-\frac{1}{10}}=\frac{1}{9}\left(\frac{1}{10}+\frac{1}{100}+\cdots\right)=\frac{1}{81}$$
As a picture:

