A poster on SE wondered how he could prove the following:

$$\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$$

Well! I never would have thought of this on my own, but I’m not really here to think up stuff like this on my own. Rather, I am here to solve the problems of confused souls, and here the solution coms from a very unexpected source.

Consider the function

$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$

So I told you it is unexpected. Out of the blue really. How, do you ask, did I come up with this? Well, it will sort of come together at the end. I saw what the product actually looked like in terms of binomial coeffiicients, and then tried to get a generating function from there. Honestly, it is the sort of trial-and-error approach that is hard to encapsulate in a well-ordered forum like this. So I beg you, just stay with me.

$f(x)$ has a Maclurin expansion as follows:

$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$

Differentiating, we get

$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$

Evaluate at $x=1/\sqrt{2}$:

$$f’\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{2}+2 = \sum_{n=0}^{\infty} \frac{2^{n}}{\displaystyle \binom{2 n}{n}} $$

Thus we have established that

$$\sum_{n=2}^{\infty} \frac{2^{n}}{\displaystyle \binom{2 n}{n}} = \frac{\pi}{2}$$

Now consider the original sum:

$$\begin{align}\sum_{n=2}^{\infty} \prod_{k=1}^n \frac{2 k}{n+k-1}&= \sum_{n=2}^{\infty}\frac{2^n n!}{n (n+1) \cdots (2 n-1)}\\ &=\sum_{n=2}^{\infty}\frac{2^n n! (n-1)!}{(2 n-1)!}\\ &= 2 \sum_{n=2}^{\infty}\frac{2^n}{\displaystyle \binom{2 n}{n}} \\&= 2 \frac{\pi}{2} \\ &= \pi \end{align} $$

The above result may also be used to prove that

$$\sum_{n=0}^{\infty }(-1)^n\ \frac{4^{n}(n!)^{2}}{(2n)!}$$

while divergent, may be assigned a finite value in the sense of a certain limiting process I outline below.

Making the substitution $x \mapsto i x$ in the expansion of $f'(x)$, I get

$$\sum_{n=0}^{\infty} (-1)^n \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n} = \frac{1}{1+x^2} – \frac{x \, \log{(x+\sqrt{1+x^2})}}{(1+x^2)^{3/2}}$$

The radius of convergence of this series is in fact $1$. Well, sort of. Obviously, at $x=1$, the series in fact diverges as demonstrated above. But the limit of the sum as $x \to 1^-$ exists and is equal to

$$\frac12 \left (1-\frac{\log{(1+\sqrt{2})}}{\sqrt{2}} \right ) \approx 0.188387$$

Dear Ron,

This is an impressive solution. Thank you very much. I have a question though; how did you go about calculating Maclaurin series for $f(x)$? Did you directly find the n-th derivative?

P.S. Are you okay with me asking question like this in a blog, or should I find the particular SE thread and comment there under your answer?

It is OK to ask me here. No, I did not find the nth derivative. I just kind of…guessed. Really, I just stumbled on it. I also knew the expansion for arcsin, so that helped, but really, I just kind of played around until I saw what I wanted. I know that is not a great answer, but it is the truth.