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

Question

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 .

Will someone please help me ?

Thanks in advance

Answer 1

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.

Answer 2

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 $\infty$.