# Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$

How does one sum the series $$S = a -\frac{2}{3}a^{3} + \frac{2 \cdot 4}{3 \cdot 5} a^{5} – \frac{ 2 \cdot 4 \cdot 6}{ 3 \cdot 5 \cdot 7}a^{7} + \cdots$$

This was asked to me by a high school student, and I am embarrassed that I couldn’t solve it. Can anyone give me a hint?!

#### Solutions Collecting From Web of "Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$"

HINT $\quad \:\;\;\rm (a^2+1) \: S’ = 1 – a \: S \;\:$ by transmuting the coefficient recurrence to a differential equation.

$\rm\;\Rightarrow\; 1 = (a^2+1) \: S’ + a \: S \; = \; f \: (f \; S)’ \;\;$ for $\rm\;\; f = (a^2+1)^{1/2}$

$\rm\displaystyle\;\Rightarrow\; S = f^{-1} \int \; f^{-1} = \frac{\sinh^{-1}(a)}{(a^2+1)^{1/2}}$

You can use the formula

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx} = \frac{2 \cdot 4 \cdot 6 \cdots 2k}{3\cdot 5 \cdots (2k+1)}$

This is called Wallis’s product.

So we have $\displaystyle S(a) = \sum_{k=0}^{\infty} (-1)^k a^{2k+1} \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx}$

Interchanging the sum and the integral

$\displaystyle S(a) = \int_{0}^{\frac{\pi}{2}}{\sum_{k=0}^{\infty}{(-1)^{k}(a\sin x)^{2k+1}} dx}$

The sum inside the integral is a geometric series of the form

$\displaystyle x – x^3 + x^5 – \cdots = x(1 – x^2 + x^4 – \cdots) = \frac{x}{1+x^2}$

Hence,

$\displaystyle S(a) = \int_{0}^{\frac{\pi}{2}}{\frac{a\sin x}{1 + (a\sin x)^2}}dx$

Now substitute $\displaystyle t = a \cos x$

The integral becomes

$\displaystyle \int_{0}^{a}{\frac{1}{1+a^2 – t^2}}dt = \frac{1}{2\sqrt{a^2+1}}\ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a} \right)$

Now $\displaystyle \ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a}\right) = \ln \left(\frac{\left(\sqrt{a^2+1}+a \right)^2}{\left(\sqrt{a^2+1}-a \right)\left(\sqrt{a^2+1}+a \right)}\right) = 2\ln \left(\sqrt{a^2+1}+a \right)$

So

$\displaystyle S(a) = \frac{1}{\sqrt{a^2+1}}\ln \left(\sqrt{a^2+1}+a \right) = \frac{\sinh^{-1}(a)}{\sqrt{a^2+1}}$

Making my comments more explicit:

Your sum of interest is

$\sum_{j=0}^\infty {(-1)^j \frac{(2j)!!}{(2j+1)!!} a^{2j+1}}$

where $(2j)!!=2\cdot 4\cdot 6\cdots (2j)$ and $(2j+1)!!=3\cdot 5\cdot 7\cdots (2j+1)$.

To simplify things a bit, we rearrange the series a bit to

$a\sum_{j=0}^\infty {\frac{(2j)!!}{(2j+1)!!}\left(-a^2\right)^j}$

The double factorials can be also expressed as

$(2j)!!=2^j j!=2^j (1)_j$

and

$(2j+1)!!=2^j \left(\frac32\right)_j$

where $(a)_j$ is a Pochhammer symbol.

Substitute both expressions into the series, and then note that the series now looks like a hypergeometric series. Now you can employ the formula here.