# 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?

• I think you mean 1/81. Commented Oct 19, 2013 at 1:36
• Whoops. Right you are. Fixed. Commented Oct 19, 2013 at 1:37
• You have an Arithmetico Geometric Series. Commented Oct 19, 2013 at 5:12

## 4 Answers

Hint:

$$f(x) = \frac{1}{1 - x} = \sum\limits_{k = 0}^{\infty} x^k$$

and so differentiating term-by-term,

$$\frac{1}{(1 - x)^2} = \sum\limits_{k = 0}^{\infty} k x^{k - 1} = \sum\limits_{n = 1}^{\infty} (n - 1) x^n$$

• Although a great hint, how can we use this to see that $f(x)$ converges to a certain limit, where in this case is $1/81$? Commented Oct 19, 2013 at 1:39
• @user67527 I don't understand your question. $f(x)$ isn't converging to a certain limit - it's just a function. Try plugging in an appropriate $x$ value.
– user61527
Commented Oct 19, 2013 at 1:40
• T. Bongers is using the geometric series summation formula (in case you missed it). Very nice construction, plug in the appropriate $x$ value in the infinite series and see what you get! Commented Oct 19, 2013 at 1:43
• IF we did not know that $\sum\limits_{n=1}^\infty (n-1)/10^n=1/81$, or in other words we were looking at $\sum\limits_{n=1}^\infty (n-1)/10^n = ?$, how could we find this $1/81$? Does that make sense? Commented Oct 19, 2013 at 1:44
• @T.Bongers: Almost at the end, we want to multiply by $x^2$, the switch from $k$ to $n$ would give $(n+1)$, not $(n-1)$. Commented Oct 19, 2013 at 1:54

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, $$\sum_{n=1}^{\infty}\frac{(n-1)}{10^n}=\frac{1}{9}\cdot\frac{1}{9}=\frac{1}{81}$$

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 $\frac1{10^2}+\frac2{10^3}+\frac3{10^4}+\frac4{10^5}+\dots=A^2=\frac1{81}$.

$$\sum\limits_{n=1}^\infty \frac{n-1}{10^n} = \sum\limits_{n=1}^\infty \sum\limits_{i=1}^{n-1} \frac{1}{10^n}= \sum\limits_{i=1}^\infty \sum\limits_{n=i+1}^\infty \frac{1}{10^{n}} = \sum\limits_{i=1}^\infty\frac{10^{-(i+1)}}{1-\frac{1}{10}}=\frac{1}{9}\left(\frac{1}{10}+\frac{1}{100}+\cdots\right)=\frac{1}{81}$$

As a picture: