How can I prove that this infinite product equals $0$?

Question

I am having trouble to prove that $$\prod_{n=1}^\infty\frac{2n-1}{2n}=0$$ I know that the sequence of partial products $(p_n)$ converges to a limit $L\ge0$, because $$p_n=\frac{(2n)!}{2^{2n}(n!)^2}$$ is decreasing and bounded below by zero.

I believe that proving that the limit equals zero would involve finding a sequence $(a_n) \to 0$ such that, eventually, $p_n \le a_n$; however, I could not find such sequence.

How can it be done?

Answer 1

An approach that does not rely on combinatorics or factorials:

$$ \left(\prod_{n=1}^\infty\frac{2n-1}{2n}\right)\left(\prod_{n=1}^\infty\frac{2n}{2n+1}\right)=\prod_{n=1}^\infty\frac{2n-1}{2n+1}=\lim_{n\to\infty} \frac1{2n+1} = 0 $$ where the $\lim$ expression arises due to telescoping.

But both of the terms on the left are positive, and it is easy to confirm that $$ \frac{2n-1}{2n}<\frac{2n}{2n+1} $$ and thus $$ a=\left(\prod_{n=1}^\infty\frac{2n-1}{2n}\right)\leq\left(\prod_{n=1}^\infty\frac{2n}{2n+1}\right)=b $$ where equality occurs if both are zero. Therefore, as $0\leq a\leq b$ and $ab=0$, we must conclude that $a=0$.

Answer 2

You may notice that: $$ p_n = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right) = \frac{1}{4^n}\binom{2n}{n} \tag{1}$$ has a square given by: $$ p_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2} $$ In particular: $$ p_n^2 \leq \frac{1}{4n}\exp\sum_{k=2}^{n}\left(\frac{1}{4(k-1)}-\frac{1}{4k}\right) \leq \frac{e^{1/4}}{4n}\tag{3} $$ and: $$ p_n \leq \frac{C}{\sqrt{n}}\tag{4} $$ with $C=\frac{1}{2}e^{1/8}$.

Answer 3

Apply Stirling's approximation $$n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ to $$\frac{(2n)!}{2^{2n}(n!)^2}.$$ Simplifying gives $$\frac{1}{\sqrt{n\pi}}\to 0.$$

Answer 4

Recall the following

Theorem: let $(\alpha_n)_{n\in \mathbb N}$ be a sequence of elements such that $\alpha_n\in [0,1)$ for every $n\in\mathbb N$. Then $$\prod_{n=1}^{+\infty}(1-\alpha_n)=0 \iff \sum_{n=1}^{+\infty}\alpha_n=+\infty$$

and just write $$\prod_{n=1}^{+\infty}\frac{2n-1}{2n}=\prod_{n=1}^{+\infty}\left(1-\frac{1}{2n}\right)$$ Since $\left(\frac{1}{2n}\right)_{n\in \mathbb N}$ is not summable, the claim follows.

Answer 5

Take the log of the left hand side.

Well, take the log of the partial products of the left hand side.

Show that the log of the partial products of the left hand side is unbounded below.

This, together with your "it actually converges", gives you the fact the partial products converge to zero.

$$\ln(\prod_{n=1}^{k}(\frac{2n-1}{2n})) = \sum_{n=1}^{k}(\ln(2n-1)-\ln{2n})$$ Now, $\ln(2n-1)-ln(2n)$ is roughly $-\frac{1}{2n}$. This is because the harmonic series (partial sums of $\frac{1}{n}$) approximates $\ln$.

$$\ln(\prod_{n=1}^{k}(\frac{2n-1}{2n})) \approx \sum_{n=1}^k(-\frac{1}{2n})$$ The right hand side is unbounded below, so the left hand side converges to zero.

All you need to do is make the $\ln(2n-1)-ln(2n) \approx -\frac{1}{2n}$ less wiggly. Stating it is less than $-\frac{1}{n}$ should be both easy to prove and more than sufficient; and we only need it "for sufficiently large n" (as we don't care about any prefix).

Answer 6

Notice that $$p_{n+1}=\frac{p_n}{2(n+1)}$$ and since $p_1=\frac{1}2$, we find that $$p_n=\frac{1}{2^n n!}$$ which tends to $0$ as $n\rightarrow\infty$.