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Can someone give some hints on how to evaluate this sum:

$$\sum_{n=0}^\infty \binom{2n+2}{n}\left(\frac{2}{3}\right)^n\left(\frac{1}{3}\right)^{n+2}$$

I came across this sum when I was solving a probability problem.

RobPratt
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Dennis
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    What was the problem? Perhaps there's an easier way to get the answer instead of reducing to an infinite sum? – Vercingetorix Nov 04 '22 at 23:19
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    $$\left(\frac23\right)^n\left(\frac13\right)^{n+2}=\frac19\left(\frac29\right)^n.$$ Not sure if that helps. – Thomas Andrews Nov 04 '22 at 23:20
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    $$\binom{2n+2}{n}=\frac{n+1}{n+2}\binom{2n+2}{n+1},$$ and you might be able to use the generating function for the central binomial coefficients to get a value for this series. – Thomas Andrews Nov 04 '22 at 23:24
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    Here it was solved for a general $K$ https://math.stackexchange.com/questions/3713734/evaluation-of-the-expression-sum-limits-n-0-infty142n-choosenp . For us, $K=2$, and the answer comes out to be $3/4$. – HackR Nov 04 '22 at 23:38
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    To add to @ThomasAndrews you may find something by simplifying the binomial coefficient using: https://math.stackexchange.com/questions/7778/proving-2n-choose-n-2n-frac-1-cdot-3-cdot-5-cdots-2n-1n – Alborz Nov 05 '22 at 00:42
  • @Vercingetorix the problem is as follows: Consider an ant walking along the positive integers. At n, the ant moves to n+1 with probability $\frac{1}{3}$ and to n−1 with probability $\frac{2}{3}$. If the ant reaches 0, it stops walking. Starting from n=3, what's the probability that the ant reaches n=5 before reaching n=0?

    This problem can be easily solved using recursion, but I was just trying to solve it in a more 'combinatoric-ish' way. But now I think that my original sum is not correct. I overcounted : p

     – Dennis Nov 05 '22 at 22:11

2 Answers2

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Looks like the expected number of visits to state $-2$, starting from state $0$, in a random walk on $\{-2,-1,0,\dots\}$ with the following transition probabilities. At state $-2$, you go to state $0$ with probability $1$; otherwise, you go up one unit with probability $2/3$ and down one unit with probability $1/3$. Let $v_i$ be this expectation, given that you start at state $i$. By first-step analysis (conditioning on the first step out of state $i$), we have $v_{-2}=1+v_0$ and $$v_i = \frac{2}{3}v_{i+1} + \frac{1}{3}v_{i-1} \quad \text{for $i > -2$}$$ Now solve this recurrence, and you will find that $v_0=3/4$.

RobPratt
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  • Good answer! How do you typically evaluate this recursion though? I'm aware of the general steps when you also have an upper bound but I'm not sure when you only have a lower bound to work with. – Vercingetorix Nov 04 '22 at 23:40
  • @Vercingetorix Generating functions are useful in cases like this. – RobPratt Nov 04 '22 at 23:42
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    Um, isn't it the expected number of times you reach $-2,$ for all time? Because it sums over all $n.$ – Thomas Andrews Nov 04 '22 at 23:53
  • @ThomasAndrews Thanks for the corrected interpretation. I didn't work out the details, just gave hints, but the original sum is $3/4$. – RobPratt Nov 05 '22 at 00:19
  • Why is this an expectation? I was thinking ${2n+2 \choose n}(2/3)^{n}(1/3)^{n+2}$ just gave the probability of reaching $-2$ in exactly $2n+2$ steps ($n$ ups and $n+2$ downs and ${2n+2 \choose n}$ ways that could happen), and as you sum over all $n$ you'd be find the probability of reaching $-2$ in any $2n+2$ steps, i.e, prob of reaching $-2$ at some point. if it was expectation then you would have a factor of $n$ there in each summand. – Vercingetorix Nov 05 '22 at 00:22
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    @Vercingetorix Because you are counting every possible round the probability on that round. So if $X_n$ is $1$ if you are at $-2$ in step $2n+2,$ and $0$ otherwise, then you are computing: $$E(X_0+X_1+X_2+\cdots).$$ To get the total probability of ever reaching $-2,$ you'd have to have $X_n$ bet $1$ if $n$ is the first round you reached $-2.$ – Thomas Andrews Nov 05 '22 at 00:32
  • Ah right. Good point. – Vercingetorix Nov 05 '22 at 00:34
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From the generating function for the central binomial coefficients we get:

$$\sum_{n=0}^{\infty}\binom{2(n+1)}{n+1}x^{n+1}=\frac{1}{\sqrt{1-4x}}-1=f(x)$$

Now, $$\binom{2n+2}{n}=\left(1-\frac{1}{n+2}\right)\binom{2n+2}{n+1}$$ and $$\left(\frac23\right)^n\left(\frac13\right)^{n+2}=\frac12\left(\frac29\right)^{n+1}.$$

So your sum can be rewritten, with $x=2/9,$ as:

$$\frac12\sum_n \binom{2n+2}{n+1}x^{n+1} -\frac1{2x}\sum_n \frac1{n+2}\binom{2n+2}{n+1}x^{n+2}.$$

So if $F$ is an antiderivative of $f$ with $F(0)=0,$ this becomes:

$$\frac12f(2/9)-\frac94F(2/9).$$

Now $f(2/9)=\frac{1}{\sqrt{1-8/9}}-1=2.$

So now we just need an antiderivative for $f.$

We can show: $$F(x)=-\frac12(1-4x)^{1/2}-x+C.$$ Then $C=1/2,$ and $F(2/9)=-\frac16-\frac29+\frac12=\frac1{9}.$

So you get a result of $\frac34.$

Thomas Andrews
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