Double factorial & roots of unity filter

Question

Original problem statement:

For any positive integer $n$, let $(2n)!!$ be the product of all positive even integers less than or equal to $2n$, By convention, $0!!=1$. For example, $6!!=6\cdot4\cdot2=48$. It can be shown that $$ \sf\sum\limits_{n = 0}^{ \infty } \frac{1}{(6n)!!} = \frac{ \sqrt{e} }{3} + \frac{2}{ 3\sqrt[4]e } \cos \theta ,\\ $$ where $0≤θ<π$. Find θ. ​

The value of $\theta$ is $\sqrt{3}\over{4}$ But my answer is different.

Here is my approach:

We can start by using the Maclaurin series expansion of $\cos(x)$:

$$\sf\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$$

Substituting x = $2/(\sqrt[4]{e})$ in the series above, we get:

$$\sf\cos \theta = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \left(\frac{2}{\sqrt[4]{e}}\right)^{2n} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \frac{4^n}{e^{n/2}}$$

Next, we can use the series expansion of the exponential function:

$$ \sf e^x = \sum_{n=0}^\infty \frac{x^n}{n!} $$

Substituting $x = 2/\sqrt{e}$ in the series above, we get:

$$ \sf e^{1/2} = \sum_{n=0}^\infty \frac{1}{n!} \left(\frac{2}{\sqrt{e}}\right)^n = \sum_{n=0}^\infty \frac{2^n}{n! e^{n/2}} $$

Multiplying both sides of the equation by $1/2$ and taking the square root, we get:

$$ \sf \sqrt{e} = \sum_{n=0}^\infty \frac{2^n}{(2n)! e^{n/2}}$$

Using this result, we can write:

$$\sf\frac{\sqrt{e}}{3} + \frac{2}{3\sqrt[4]{e}} \cos \theta = \frac{1}{3} \sum_{n=0}^\infty \frac{2^n}{(2n)! e^{n/2}} + \frac{2}{3} \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \frac{4^n}{e^{n/2}} $$

Now we can combine the two series using the formula for the cosine series and get:

$$ \sf\frac{\sqrt{e}}{3} + \frac{2}{3\sqrt[4]{e}} \cos \theta = \frac{1}{3} \sum_{n=0}^\infty \frac{2^n}{(2n)! e^{n/2}} + \frac{1}{3} \sum_{n=0}^\infty \frac{2^n}{(2n)! e^{n/2}} \cos\left(\frac{n\pi}{2}\right) - \frac{1}{3} \sum_{n=0}^\infty \frac{2^n}{(2n)! e^{n/2}} \sin\left(\frac{n\pi}{2}\right) $$

Since the series for $\sin(nπ/2)$ is zero for even n and $(-1)^{(n-1)/2}$ for odd $n$, we can simplify the expression above to:

$$ \sf\frac{\sqrt{e}}{3} + \frac{2}{3\sqrt[4]{e}} \cos \theta = \frac{1}{3} \sum_{n=0}^\infty \frac{2^n}{(2n)! e^{n/2}} \left(1 + (-1)^{n/2} \cos\left(\frac{n\pi}{2}\right) \right)$$

To obtain the desired form of the series, we just need to adjust the lower limit of the sum to take into account the alternating factor, which gives:

$$\sf\frac{\sqrt{e}}{3} + \frac{2}{3\sqrt[4]{e}} \cos \theta = \frac{1}{3} \sum_{n=0}^\infty \frac{1}{(2n)!} \frac{2^{2n}}{e^{n}} \left(1 + \cos\left(\frac{n\pi}{2}\right) \right)$$

Comparing this expression to the given series, we see that:

$$\sf\frac{1}{(6n)!!} = \frac{1}{(2n)!} \frac{2^{2n}}{e^{n}} \left(1 + \cos\left(\frac{n\pi}{2}\right) \right)$$

Therefore, we can write the given series in the desired form as:

$$ \sf\sum_{n = 0}^{ \infty } \frac{1}{(6n)!!} = \frac{1}{3} \sum_{n=0}^\infty \frac{1}{(2n)!} \frac{2^{2n}}{e^{n}} \left(1 + \cos\left(\frac{n\pi}{2}\right) \right) $$

Finally, we can see that the value of θ is zero, since the cosine term has a maximum value of $1$ and is positive for all $n$. Therefore, the expression simplifies to:

$$ \sf\sum_{n = 0}^{ \infty } \frac{1}{(6n)!!} = \frac{1}{3} \sum_{n=0}^\infty \frac{1}{(2n)!} \frac{2^{2n}}{e^{n}} \left(2 \right) = \sum_{n=0}^\infty \frac{1}{n!} \frac{1}{e^{3n/2}} $$

Answer 1

For problems like this one it is usually best to perform the computation directly, rather than working backwards.

To start, let $S$ be your sum, and note that $(6n)!!=2^{3n}(3n)!$, so

$$S=\sum_{n=0}^\infty\frac{1}{(6n)!!}=\sum_{n=0}^\infty \frac{1}{2^{3n}(3n)!}$$

So $S$ sums every third term of the sequence $\frac{1}{2^n n!}$, or equivalently

$$S=\sum_{n=0}^\infty\frac{a_n}{2^n n!}$$

where $(a_n)_n$ is a sequence defined as $a_n=1$ if $3|n$ and $a_n=0$ otherwise. We may explicitly write $a_n$ as a linear combination of powers of the third roots of unity as follows: $$a_n=\frac{1}{3}(1+\xi^n+\xi^{2n})$$

where $\xi=e^{2i\pi /3}$ is a primitive third root of unity. Note that $\xi=\frac{-1+i\sqrt3}{2}$ and $\xi^2=\frac{-1-i\sqrt3}{2}$. Therefore, using the series expansion of the exponential function $e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$, and the complex identity $\cos(z)=\frac12 (e^{iz}+e^{-iz})$, we may simplify

$$\begin{split} S&=\sum_{n=0}^\infty\frac{1}{3}\cdot\frac{1+\xi^n+\xi^{2n} }{2^n n!}\\ &=\frac13\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{1}{2}\right)^n+ \frac13\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{\xi}{2}\right)^n+ \frac13\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{\xi^2}{2}\right)^n\\ &=\frac13 e^{1/2}+\frac13e^{\xi/2}+\frac13 e^{\xi^2/2}\\ &=\frac13 e^{1/2}+\frac13e^{-1/4}(e^{i\sqrt3/4}+e^{-i\sqrt3/4})\\ &=\frac13 e^{1/2}+\frac23e^{-1/4}\cos\left(\frac{\sqrt3}4\right) \end{split}$$ In other words, $\theta=\frac{\sqrt3}4$, as desired.