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I have to prove that the following product converges:
$\prod\limits_{k=1}^n \frac{2k-1}{2k}$

I've seen convergence of sums before, but this is new to me. In fact, I had to look up the $\Pi$ notation. I figure I can't use the usual approaches here and since it was not introduced in my course, I'm pretty sure there has to be a simple trick to it, right?

Any help is appreciated.

John
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  • Since the upper limit is $k=n$ rather than $\infty$, there is no question of convergence -- it is just an ordinary product of finitely many factors. – hmakholm left over Monica Nov 21 '15 at 19:59
  • Should there be a limit somewhere? – Mankind Nov 21 '15 at 20:00
  • This product does not converge (assuming you mean to take $n \to \infty$). It goes to $0$, and the standard convention is that if a product goes to $0$ then it diverges. The convention is made this way so that, for example, a product of positive reals converges iff its logarithm converges as a sum. – Qiaochu Yuan Nov 21 '15 at 20:03
  • Well, it basically reads "prove that it converges", so unless there's a typo in the assignment, it's reasonable to assume that it converges, right? – John Nov 21 '15 at 20:05
  • @John: the correct statement is that the sequence $a_n = \prod_{k=1}^n \frac{2k-1}{2k}$ converges (to $0$), as in HowDoIMath's answer. But there is a different convention for what it means for an infinite product to converge, and with respect to this convention, the infinite product $\prod_{k=1}^{\infty} \frac{2k-1}{2k}$ "diverges to $0$." – Qiaochu Yuan Nov 21 '15 at 20:08

3 Answers3

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Define $a_n=\prod\limits_{k=1}^n \frac{2k-1}{2k}$. I presume that the problem is to show that the sequence $\{a_n\}$ converges. Look at this:

$$a_n=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}.$$

Notice how each consecutive factor is a number less than $1$. Hence $a_n>a_{n+1}$ for all $n$. Since also $a_n>0$ for all $n$, the sequence $\{a_n\}$ is monotone and bounded, thus convergent.

Mankind
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A calculation to exactly prove the limit, using Stirling's approximation: $$ \left(\frac{n}{e}\right)^n\sqrt{2\pi n}<n!<\left(\frac{n}{e}\right)^n\sqrt{2\pi n}e^{1/(12n)} $$ Then: $$ \prod_{i=1}^n\frac{2i-1}{2i}=\prod_{i=1}^n\frac{(2i-1)\cdot 2i}{2^2i^2}=\frac{(2n)!}{2^{2n}(n!)^2} $$ The lower bound: $$ \frac{(2n)!}{2^{2n}(n!)^2}\geq \frac{\displaystyle \left(\frac{2n}{e}\right)^{2n}\sqrt{2\pi (2n)}}{2^{2n}\displaystyle\left(\left(\frac{n}{e}\right)^n\sqrt{2\pi n}e^{1/(12n)}\right)^2}=\frac{\exp(-1/(6n))}{\sqrt{n\pi}} $$ The upper bound: $$ \frac{(2n)!}{2^{2n}(n!)^2}\leq\frac{\exp(1/(24n))}{\sqrt{n\pi}} $$ Thus: $$ \frac{\exp(-1/(6n))}{\sqrt{n\pi}}\leq \prod_{i=1}^n\frac{2i-1}{2i}\leq \frac{\exp(1/(24n))}{\sqrt{n\pi}} $$ Since both sides converge to $0$, we are done.

asomog
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  • This is way too much effort. Just take the logarithm and bound $\log \left( 1 - \frac{1}{2i} \right)$ by $- \frac{1}{2i}$. – Qiaochu Yuan Nov 21 '15 at 21:46
  • Yes, but it also gives asymptotics on the speed of convergence. ( And I personally like using Stirling) – asomog Nov 21 '15 at 21:57
  • The speed of convergence is clear from taking the logarithm; it grows like $\frac{1}{\sqrt{n}}$ up to a multiplicative constant. – Qiaochu Yuan Nov 21 '15 at 21:59
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Set $u_n=\prod_{k=1}^n \frac{2k-1}{2k}$ which is a strictly positive sequence. Now set $v_n=log(u_n)$ and $w_n=v_{n}-v_{n-1} = log(2n-1)-log(2n)$. Then $\sum -w_n$ is a positive series.

$-w_n = log(1+\frac{1}{2n-1}) = \frac{1}{2n-1} + O(\frac{1}{n^2})$ thus the sum $v_n$ is divergent to $+\infty$ because the harmonic series has limit $+\infty$.

That implies $v_n \rightarrow -\infty$ and $u_n = exp(v_n) \rightarrow 0$.

Akos' proof uses the Stirling formula. This is neat but this formula (minus the $\sqrt{2 \pi}$ constant) is proven using series and logs in the same way as here, so this is hiding the difficulty. Moreover, determining the $\sqrt{2 \pi}$ requires using the Wallis integrals which are somehow similar to $\prod_{k=1}^n \frac{2k-1}{2k}$ so this is too heavy of a result for such a simple fact to prove.

Béranger Seguin
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