Calculate the sum of the following series

Question

I would like to find the sum of the following series:

$$\sum_{n=0}^{\infty} \binom{2n+1}{n+1} (p(1-p))^n$$

where $0<p<\frac{1}{2}$. I put it on Wolfram and it said that this series converges to $$\frac{\sqrt{(2p-1)^2}-1}{2p(p-1)\sqrt{(2p-1)^2}}$$

I initially thought it was a Geometric progression, but turns out its not.

I am guessing it could work with binomial series, where we have $\sum_{k=0}^{\infty}\binom{\alpha}{k}x^k=(1+x)^{\alpha}$. But here, my $\alpha$ depends on $k$, so I'm slightly stuck. I am aware that finding the sum of a series is hard and am unsure of how else to go about finding the sum of this series. Any help would be appreciated!

Answer 1

The series representation \begin{align*} \color{blue}{\sum_{n=0}^{\infty}\binom{2n+1}{n+1}\left(p(1-p)\right)^n}\tag{1} \end{align*} is nice and instructive.

WA: We start with the result provided by Wolfram Alpha. We obtain for $0<p<\frac{1}{2}$ \begin{align*} \color{blue}{\frac{\sqrt{(2p-1)^2}-1}{2p(p-1)\sqrt{(2p-1)^2}}} &=\frac{1-|2p-1|}{2p(1-p)|2p-1|}\\ &=\frac{1-(1-2p)}{2p(1-p)(1-2p)}\\ &=\frac{1}{(1-p)(1-2p)}\\ &=\frac{1}{p}\left(\frac{1}{1-2p}-\frac{1}{1-p}\right)\\ &=\frac{1}{p}\left(\sum_{j=0}^{\infty}\left(2p\right)^j-\sum_{j=0}^{\infty}p^j\right)\\ &=\frac{1}{p}\sum_{j=1}^{\infty}\left(2^j-1\right)p^j\\ &\,\,\color{blue}{=\sum_{j=0}^\infty\left(2^{j+1}-1\right)p^j}\tag{2}\\ &=1+3p+7p^2+15p^3+63p^4+\cdots \end{align*}

So, according to WA we see that a series expansion of (1) at $p=0$ gives the nice and simple difference of geometric series in (2).

But of course, this has to be shown. We use the coefficient of operator $[p^t]$ to denote the coefficient of $p^t$ of a series.

Coefficient extraction:

We obtain from (1) \begin{align*} \color{blue}{[p^t]}&\color{blue}{\sum_{n=0}^{\infty}\binom{2n+1}{n+1}\left(p(1-p)\right)^n}\\ &=\sum_{n=0}^t\binom{2n+1}{n+1}[p^t]\left(p(1-p)\right)^n\tag{3.1}\\ &=\sum_{n=0}^t\binom{2n+1}{n+1}[p^{t-n}](1-p)^n\tag{3.2}\\ &\,\,\color{blue}{=\sum_{n=0}^t\binom{2n+1}{n+1}\binom{n}{t-n}(-1)^{t-n}}\tag{3.3} \end{align*}

Comment:

• In (3.1) we use the linearity of the coefficient of operator and set the upper limit of the series to $t$ since other values of $n$ do not contribute to the coefficient of $p^t$.

• In (3.2) we use the identity $[p^{t-n}]A(p)=[p^t]p^nA(p)$.

• In (3.3) we select the coefficient of $p^{t-n}$.

Binomial identity:

Since the coefficient of $p^t$ in (2) is according to WA equal to (3.3) we have finally to show the nice binomial identity: \begin{align*} \color{blue}{\sum_{n=0}^t\binom{2n+1}{n+1}\binom{n}{t-n}(-1)^{t-n}=2^{t+1}-1\qquad\qquad t\geq 0}\tag{4} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{n=0}^t}&\color{blue}{\binom{2n+1}{n+1}\binom{n}{t-n}(-1)^{t-n}}\\ &=\sum_{n=0}^t\binom{t+1}{n+1}\binom{2n+1}{t+1}(-1)^{t-n}\tag{4.1}\\ &=\sum_{n=1}^{t+1}\binom{t+1}{n}\binom{2n-1}{t+1}(-1)^{t+1-n}\tag{4.2}\\ &=\sum_{n=1}^{t+1}\binom{t+1}{n}[z^{t+1}](1+z)^{2n-1}(-1)^{t+1-n}\tag{4.3}\\ &=(-1)^{t+1}[z^{t+1}](1+z)^{-1}\sum_{n=1}^{t+1}\binom{t+1}{n}\left(-(1+z)^2\right)^n\tag{4.4}\\ &=(-1)^{t+1}[z^{t+1}](1+z)^{-1}\left(\left(1-(1+z)^2\right)^{t+1}-1\right)\tag{4.5}\\ &=(-1)^{t+1}[z^{t+1}](1+z)^{-1}\left(\left((-2z-z^2\right)^{t+1}-1\right)\\ &=[z^0](1+z)^{-1}(2+z)^{t+1}-(-1)^{t+1}[z^{t+1}](1+z)^{-1}\\ &=[z^0]\sum_{j=0}^{t+1}\binom{t+1}{j}2^jz^{t+1-j}(1+z)^{-1}-1\tag{4.6}\\ &\,\,\color{blue}{=2^{t+1}-1}\tag{4.7} \end{align*} and the claim follows.

Comment:

• In (4.1) we use the binomial identity $\binom{2n+1}{n+1}\binom{n}{t-n}=\binom{t+1}{n+1}\binom{2n+1}{t+1}$.

• In (4.2) we shift the index to start with $n=1$.

• In (4.3) we use the representation $[z^t](1+z)^n=\binom{n}{t}$.

• In (4.4) we factor out terms which do not depend on $n$.

• In (4.5) we apply the binomial theorem.

• In (4.6) we use $[z^{t+1}](1+z)^{-1}=[z^{t+1}]\left(1-z+z^2-z^3+\cdots\right)=(-1)^{t+1}$ and we also apply the binomial theorem.

• In (4.7) we observe that only $j=t+1$ contributes to the coefficient of $z^0$.