Summing a series having a geometric component

Question

If $$\alpha=\frac {5}{2!3}+\frac {5\cdot 7}{3!3^2}+\frac {5\cdot 7\cdot 9}{4!3^3}+\dots $$ Find the value of $\alpha^2+4\alpha $.

The possible options are:

1. 21

2. 23

3. 25

4. 27

I think $$\alpha=\sum \frac {(2n+4)!}{2^{n+2}(n+1)!(n+2)!3^{n+1}}$$ but cannot think of what else to do.

Answer 1

Consider the binomial expansion of $$(1-\frac 23)^{-\frac 32}$$

This has value $3\sqrt{3}$ and when expanded gives the sequence $$1+1+\frac{5}{2!3}+\frac{5\cdot7}{3!3^2}+...$$

So $$\alpha=3\sqrt{3}-2$$ which gives the required expression the value of $23$

Answer 2

Indeed $$\alpha=\dfrac16\sum_{n=2}^\infty\dfrac{(2n+1)!}{(n!)^2}\left(\dfrac16\right)^n$$ one may construct a binomial series and show that with the generating function $$\alpha(x)=-\dfrac16-x+\dfrac16\dfrac{1}{\sqrt{1-4x}^3}$$ then $\alpha=\alpha(\dfrac16)$.

Answer 3

You have $$\alpha = \sum_{j=1}^\infty \frac{(2j+3)!! }{(j+1)! 3^{j+1}},$$ where $p!!$ is the double factorial of the nonnegative integer $p$: $$p!! = \begin{cases} p \cdot (p-2)\cdot (p-4) \cdots 4\cdot 2 & p\ \text{even}\\ p \cdot (p-2)\cdot (p-4) \cdots 5\cdot 3\cdot 1 & p\ \text{odd,} \end{cases}$$ or, recursively, $$\begin{cases} 0!! = 1!! = 1\\ p!! = p(p-2)!! & \forall p\geq 2. \end{cases}$$ We may rearrange the indices, setting $n=j+1$, to find $$\alpha = \sum_{n=2}^\infty \frac{(2n+1)!!}{n!3^{n}} = -2 + \sum_{n=0}^\infty \frac{(2n+1)!!}{n!3^n} =: \beta -2 .$$ By using the basic identities $$(2n+1)!! = \frac{(2n+1)!}{(2n)!!}, \qquad (2n)!! = n!2^n,$$ we may simplify $$\beta = \sum_{n=0}^\infty \frac{(2n+1)!}{n!^2 6^n}. $$ Now notice that $$\alpha^2 + 4\alpha = 4 + \beta^2 -4\beta -8 + 4\beta = \beta^2 -4 .$$ We can calculate $\beta^2$ as a product of two infinite sums by virtue of Cauchy’s theorem: $$\begin{split} \beta^2 = \beta\cdot\beta &= \sum_{n=0}^\infty \sum_{k=0}^n \frac{(2k+1)!(2n-2k+1)!}{k!^2(n-k)!^2 6^n} \\&= \sum_{n=0}^\infty \frac{1}{n!^26^n}\sum_{k=0}^n \binom n k^2(2k+1)!(2n-2k+1)!. \end{split}$$ By employing the identity $$ \sum_{k=0}^n \binom n k^2(2k+1)!(2n-2k+1)! = 2^{2n-1} n! (n+2)! = 2^{2n-1} (n+1)(n+2) n!^2, \tag{$\ast$}$$ which may be settled by induction (EDIT: proven below), we obtain $$\begin{split} \beta^2 &= \sum_{n=0}^\infty \frac{2^{2n-1} (n+1)(n+2) }{2^n3^n} \\&=\frac 12 \sum_{n=0}^\infty \left(\frac 23\right)^n (n^2 +3n + 2) \\ &= \frac 12\lim_{m\to\infty} \left[\sum_{n=0}^m n^2 \left(\frac 23\right)^n + 3 \sum_{n=0}^m n \left(\frac 23\right)^n + 2 \sum_{n=0}^m \left(\frac 23\right)^n\right], \end{split}$$ which can be shown to converge to $27$. Hence $$\alpha^2 + 4\alpha =23. $$

---

ADDENDUM. For the sake of completeness, I will prove the identity $(\ast)$ by induction. The base case ($n=1$) is immediate: $$\color{grey}{\binom 1 0^2} \cdot \color{grey}{1!} \cdot 3! + \color{grey}{\binom 1 1^2 }\cdot 3! \cdot\color{grey}{ 1!} = 12 = 2^{\color{grey}1} \cdot \color{grey}{1!} \cdot 3!. $$ Now suppose the statement holds for $n$: $$ \color{darkorange}{\gamma_{n}} := \sum_{k=0}^{n} \binom{n}{k}^2 (2k+1)! (2n-2k+1)! = \color{darkorange}{2^{2n-1} n!(n+2)!} . $$ We need to show $$ \color{limegreen}{\gamma_{n+1}} := \sum_{k=0}^{n+1} \binom{n+1}{k}^2 (2k+1)! (2n-2k+3)! \overset{(?)}{=} \color{limegreen}{2^{2n+1} (n+1)!(n+3)!}. \tag{$\circ$}$$ The idea is to somehow massage the LHS of $(\circ)$ so that the original sum $\color{darkorange}{\gamma_n}$, or a related sum, shows up, and then to use the inductive hypothesis. To do so, we may first reduce the binomial coefficient $\binom {n+1}{k}$ to binomial coefficients which only have $n$ on top, by using Pascal's rule: for all $k$ such that $1 \leq k \leq n$, $$ \binom {n+1}k = \binom n{k-1} + \binom nk.$$ In order to substitute this in, we need to make sure the running index goes from $1$ to $n$, by extracting the first term in advance; we may also extract the $(n+1)$-th term in order to normalize the sum. $$\begin{split} \color{limegreen}{\gamma_{n+1}} &= (2n+3)! + \sum_{k=1}^{n+1} \binom{n+1}{k}^2 (2k+1)! (2n-2k+3)! \\ &= 2(2n+3)! + \sum_{k=1}^{n} \binom{n+1}{k}^2 (2k+1)! (2n-2k+3)! \\ &= 2(2n+3)! + \sum_{k=1}^{n} \left[\binom{n}{k-1}^2 + \binom nk^2 + 2\binom n{k-1} \binom nk\right] (2k+1)! (2n-2k+3)! \\ &= 2(2n+3)! + \sum_{k=1}^{n} \binom{n}{k-1}^2 (2k+1)! (2n-2k+3)! \\ &\quad + \sum_{k=1}^{n} \binom nk^2 (2k+1)! (2n-2k+3)! + 2 \sum_{k=1}^{n}\binom n{k-1} \binom nk (2k+1)! (2n-2k+3)!. \end{split}$$ Let us work on these last three sums. Notice that, setting $j=k-1$, the first sum becomes $$\sum_{k=1}^{n} \binom{n}{k-1}^2 (2k+1)! (2n-2k+3)! = - (2n+3)! + \sum_{j=0}^n \binom{n}{j}^2 (2j+3)! (2n-2j+1)!, $$ and extending the range of the second sum yields $$\sum_{k=1}^{n} \binom nk^2 (2k+1)! (2n-2k+3)! = - (2n+3)! + \sum_{k=0}^{n} \binom nk^2 (2k+1)! (2n-2k+3)!. $$ The two boundary terms $-(2n+3)!$ cancel out with those that were accumulated earlier. Also, the product of two binomial coefficients in the third sum can be rearranged as $$\binom n{k-1} \binom nk = \frac{n!}{(k-1)!(n-k+1)!}\frac{n!}{k!(n-k)!} = \frac{n!^2}{k!^2(n-k)!^2} \frac k {n-k+1}. $$ All in all, $$\begin{split} \color{limegreen}{\gamma_{n+1}} &= \sum_{k=0}^n \binom{n}{k}^2 \color{brown}{(2k+3)(2k+2)}(2k+1)! (2n-2k+1)! \\ &\quad + \sum_{k=0}^{n} \binom nk^2 (2k+1)! \color{brown}{(2n-2k+3)(2n-2k+2)}(2n-2k+1)! \\ &\quad + \sum_{k=0}^{n} \binom nk^2 \color{brown}{\frac {2k} {n-k+1}} (2k+1)! \color{brown}{(2n-2k+3)(2n-2k+2)}(2n-2k+1)! \\ &= \sum_{k=0}^{n} \binom nk^2 (2k+1)!(2n-2k+1)! \color{brown}{\left[4n^2 + 10n + 6k + 12 \right] }\\ &= \color{brown}{(4n^2 + 10n + 12) }\color{darkorange}{\gamma_n} + \color{brown}{12 }\sum_{k=0}^n \color{brown}k \binom nk^2 (2k+1)!(2n-2k+1)!, \end{split}$$ where I have highlighted in brown the "extra factors". The last sum is identical in form to the original sum, but has an extra $\color{brown}k$ in the argument. By a similar argument to Gauss's proof of the sum of the first $n$ natural numbers, employing the symmetry property $$\binom nk^2 (2k+1)!(2n-2k+1)! = \binom n{n-k}^2 (2(n-k)+1)!(2n-2(n-k)+1)! ,$$ we may write $$\begin{split} 2\sum_{k=0}^n \color{blue}k \binom nk^2 (2k+1)!(2n-2k+1)! &= (\color{blue}0 + \color{blue}n) \binom n0^2 1!(2n+1)! \\ &\quad + (\color{blue}1 + \color{blue}{(n-1)}) \binom n1^2 3!(2n-1)! \\ &\quad + (\color{blue}2 + \color{blue}{(n-2)}) \binom n2^2 5!(2n-3)! + \cdots \\ &= \color{blue}n\sum_{k=0}^n \binom nk^2 (2k+1)!(2n-2k+1)! =\color{blue} n\color{darkorange}{\gamma_n}. \end{split}$$ In conclusion, by the inductive hypothesis, $$\begin{split} \color{limegreen}{\gamma_{n+1} } &= (4n^2 + 10n + 12) \color{darkorange}{\gamma_n} + 6n\color{darkorange}{\gamma_n} \\ &= 2^2(n+1)(n+3) {2^{2n-1} n!(n+2)!} = \color{limegreen}{2^{2n+1} (n+1)!(n+3)!} \end{split} $$ which is $(\circ)$, as we wanted.

Answer 4

So we have that $$\alpha=\sum_{n=0} \frac{(5+2n)!!}{3^{n+2}(2+n)!}$$ Using the fact that $(2n)!=2^n(2n-1)!!(n!)$, we have that $$\alpha=\sum_{n=0} \frac{(2n+6)!}{2^{n+3}3^{n+2}(n+2)!(n+3)!}$$ $$\alpha=\frac{1}{2}\sum_{n=0} \binom{2n+6}{n+3}\left(\frac{1}{6}\right)^{n+2}(n+3)$$ Using the generating function of central binomial coefficients, we know $$\sum_{n=0}^\infty \binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$$ Define the function $$f(x)=\sum_{n=0}^{\infty}\binom{2n+6}{n+3}x^{n+3}=\frac{1}{\sqrt{1-4x}}-1-2x-6x^2$$ We know $$f'(x)=\sum_{n=0}^\infty \binom{2n+6}{n+3}x^{n+2}(n+3)=\frac{2}{\left(\sqrt{1-4x}\right)^3}-2-12x$$ Hence, $$\alpha=\frac{1}{2}f'\left(\frac{1}{6}\right)=3\sqrt{3}-2$$ We can then deduce that $\alpha$ is a root of $(x+2)^2=27$, or $x^2+4x-23=0$. Hence, $\alpha^2+4\alpha=\boxed{23}$