# Double factorial & roots of unity filter

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:

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

• What? A lot of your substitutions/evaluates seems incorrect to me. For example, substituting $x=2/\sqrt{e}$ into $e^x$ gives $e^{2/\sqrt{e}}$ not $e^{1/2}$. You rarely want to substitute weird values like $2/e^{1/4}$ into $cos(x)$ since once you have powers of $e$ inside the summation, it’s hard to get rid of them.
– Eric
Commented May 1, 2023 at 5:06
• A standard trick when you want every kth term of a series is to plug in roots of unity and then sum it up. In this case, you may want to expand and add $cos(i)$, $cos(i \omega)$ and $cos(i \omega^2)$ where $\omega^3=1$.
– Eric
Commented May 1, 2023 at 5:09
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. You can also upvote answers that you find helpful. Commented May 8, 2023 at 18:17

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.