# Evaluate $\lim_{n\to\infty}\prod_{k=1}^{n}\frac{2k}{2k-1}$

How can I calculate the following limit:

$$\lim_{n\to \infty} \frac{2\cdot 4 \cdots (2n)}{1\cdot 3 \cdot 5 \cdots (2n-1)}$$ without using the root test or the ratio test for convergence?

I have tried finding an upper and lower bounds on this expression, but it gives me nothing since I can't find bounds that will be "close" enough to one another. I have also tried using the fact that: $2\cdot 4 \cdot...\cdot (2n)=2^n n!$ and $1\cdot 3 \cdot 5 \cdot...\cdot (2n-1) =2^n (n-0.5)!$ but it also gives me nothing .

Your product is $$\left(1+\dfrac11\right)\left(1+\dfrac13\right)\left(1+\dfrac15\right)\cdots \left(1+\dfrac1{2n-1}\right)$$ An infinite product $\lim_{n \to \infty} \left(1+a_n\right)$ converges to a non-zero number iff one of the $\sum_{n \to \infty} \vert a_n \vert$ converges. Conclude what you want from this.
Notice that $$\dfrac12 \cdot \dfrac34 \cdot \dfrac56 \cdots \dfrac{2n-1}{2n} = \left(1 - \dfrac12\right)\left(1 - \dfrac14\right)\left(1 - \dfrac16\right)\cdots\left(1 - \dfrac1{2n}\right)$$ Since $$\dfrac12 + \dfrac14 + \dfrac16 + \cdots + \dfrac1{2n} + \cdots$$ diverges, the infinite product goes to $$0$$.
So its limit goes to $$\infty$$.
• Divergence to $0$ is an abuse of terminology, which I have mentioned in one of my previous avatars on math stackexchange. – Adhvaitha Nov 22 '14 at 15:26
• @GEdgar As I have said before, it is an abuse of terminology. For instance, see page 141 of Stein and Shakarchi, where the theorem on a version of infinite product states "... converges to zero...". Also, for instance, consider the infinite product $a_n = (1-1/2)(1-1/3)(1-1/4)\cdots (1-/n)$. $a_n$ is nothing but $1/n$ and saying $a_n$ converges $0$ is correct as opposed to saying $a_n$ diverges to $0$. Divergence to zero was introduced by someone just for the reason of appearing cool before lesser mortals. – Adhvaitha Nov 22 '14 at 17:49
• In $a_n = (1-1/2)(1-1/3)(1-1/4)\cdots (1-1/n)$, the standard terminology says the sequence $a_n$ converges to zero, but the corresponding infinite product diverges to zero. You can argue with the standard terminology, but I don't think it will get you very far. Those who created this deemed it desirable that, in an infinite product, if it "converges" to zero, then one of the factors is zero. So they fixed the terminology to make this work. And this is what is useful in complex analysis. – GEdgar Nov 22 '14 at 20:48