How to calculate the sum of an infinite series, such as $\frac{n(n-1)}{2^n}$?

Question

This problem appears in Girls in Math at Yale, a high school competition which I am prepping for:

"Marie repeatedly flips a fair coin and stops after she gets tails for the second time. What is the expected number of times Marie flips the coin?"

I deduced that for there to be n flips, we have to add $\dfrac{n(n-1)}{2^n}$ to get the

I have also encountered a few similar problems like dice problems, where I have to calculate things like $\dfrac{1\cdot1}{6} + \dfrac{2\cdot1}{36} + \dfrac{3\cdot1}{216} \cdots$

So how should I approach this?

Answer 1

Basic approach. Summations of this kind can often be evaluated by differentiating a geometric sum, for which you may already know a formula. For example, let

$$ S(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x} $$

If we differentiate this with respect to $x$, we get

$$ S'(x) = \sum_{k=0}^\infty kx^{k-1} = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2} $$

which might be challenging to obtain otherwise. We can then multiply both sides by $x$ to get

$$ xS'(x) = \sum_{k=0}^\infty kx^k = \frac{x}{(1-x)^2} $$

Differentiating $S'(x)$ to get $S''(x)$, multiplying by whatever factor of $x$ you need, and then evaluating at $x = 1/2$ or $1/6$ or whatever should get you to where you need to go.

Answer 2

The problem you mention can be easily solved using probability distributions. In fact, if you take 'success' as getting tails, the probability of success is $\dfrac{1}{2}$.

As the game is set to stop when there are two tails (successes), the number of failures (heads) before the game ends is modeled by a Negative Binomial distribution with parameters $p=\dfrac{1}{2}$ and $r=2$. Its mean is $$\dfrac{r(1-p)}{p}=2$$

We expect $2$ failures before getting two successes, and so $4$ flips in total. No infinite sums needed.

Answer 3

It’s overkill for this particular problem, but I give a fully general (finite) formula for evaluating such sums with a worked example in my answer here.

In the case at hand, $$2\sum_{\text{n}\ =\ 0,\ 1,\ \dots}\binom{\text{n}}{\textbf{2}}\left(\tfrac{1}{2}\right)^{\text{n}}\ =\ 2\cdot\frac{\left(\tfrac{1}{2}\right)^{\textbf{2}}}{\left(1-\tfrac{1}{2}\right)^{\textbf{2}+1}}\ =\ 4\text{.}$$

Answer 4

$$\begin{aligned} S=\sum_{n=0}^\infty(n^2-n)2^{-n}&=\left[\sum_{n=0}^\infty(n^2-n)x^{-n}\right]_{x=2}\\ &=\left[\sum_{n=0}^\infty n^2x^{-n}-\sum_{n=0}^\infty nx^{-n}\right]_{x=2}\\ &=\left[x\frac{\text d}{\text dx}\left(x\frac{\text d}{\text dx}\sum_{n=0}^\infty x^{-n}\right)+x\frac{\text d}{\text dx}\sum_{n=0}^\infty x^{-n}\right]_{x=2}=[S_1(x)+S_2(x)]_{x=2} \end{aligned} $$ First we calculate the $2^{\text{nd}}$ term and then the $1^{\text{st}}$ one with the previous: $$ \begin{aligned} &S_2(x)=x\frac{\text d}{\text dx}\sum_{n=0}^\infty x^{-n}=x\frac{\text d}{\text dx}\left(\frac{x}{x-1}\right)=-\frac{x}{(x-1)^2}\\ &S_1(x)=x\frac{\text d}{\text dx}\left(-\frac{x}{(x-1)^2}\right)=\frac{x(x+1)}{(x-1)^3} \end{aligned} $$ Finally, we have $$S=\left[\frac{x(x+1)}{(x-1)^3}-\frac{x}{(x-1)^2}\right]_{x=2}=4$$ The cool thing about this is that now you can compute the general series for $\forall x$ as long as the summation converges ($|x|>1$).