Evaluate the limit $\lim_{n\to\infty}\prod_{r=1}^n\frac{4r}{4r+3}$

Question

Evaluate the limit $$\lim_{n\to\infty}\prod_{r=1}^n\frac{4r}{4r+3}$$

My Attempt:

I am trying to do by Squeeze theorem and then idea of telescopic product but not able to get anywhere.

Using $A.M\geq G.M$ we get $$\frac{2r+1+2r+2}{2}\geq \sqrt{(2r+1)(2r+2)}$$ $$\frac{4r+3}{4r}\geq \frac{\sqrt{(2r+1)(2r+2)}}{2r}$$ $$\frac{4r}{4r+3}\leq \frac{2r}{\sqrt{(2r+1)(2r+2)}}<\frac{2r+1}{\sqrt{(2r+1)(2r+2)}}$$ $$\frac{4r}{4r+3}< \sqrt{\frac{2r+1}{2r+2}}$$

Some clever manipulation or trick is required here

There is something here though Limit as $n\to+\infty$ of $\prod_{k=1}^{n} \frac{2k}{2k+1}$ but no general method as such

Answer 1

Using $$\log(x + 3) - \log x \geq \frac{3}{x + 3}.$$ Thus denoting $P_n := \prod_{1 \leq r \leq n} \frac{4r + 3}{4r},$ we obtain $$\log P_n \geq \sum_{1 \leq r \leq n} \frac{3}{4r + 3} \to \infty,$$ which in turn yields the convergence to $0$ of your initial product. I hope this helps. :)

Answer 2

You can show, that (this holds for $r \in (-\frac{8}{9},-1)\cup( -\frac{3}{4},\infty)$ ) $$\frac{4r}{4r+3}<\frac{4r+3}{4r+4}$$ with $$0<\frac{4r}{4r+3}, r>0$$ and $$\prod^{n}_{k=1}(\frac{4r}{4r+3})(\frac{4r+3}{4r+4})=\frac{4}{4n+4}=\frac{1}{n+1}$$ since it is a telescopic product, we get that: $$0<\prod^{n}_{k=1}(\frac{4r}{4r+3})^2<\prod^{n}_{k=1}(\frac{4r}{4r+3})(\frac{4r+3}{4r+4})=\frac{1}{n+1}\longrightarrow 0$$ thus, $$\lim_{n\to\infty}\prod_{r=1}^n\frac{4r}{4r+3}=0$$

Answer 3

A straight forward method that uses Stirling's approximation.

\begin{align} P_{n} &= \prod_{k=1}^{n} \left(\frac{4 \, k}{4 k + 3} \right) = \frac{4^n \, \Gamma(n+1)}{\prod\limits_{k=1}^{n} (4 k +3)} = \Gamma\left(\frac{7}{4}\right) \, \frac{\Gamma(n+1)}{\Gamma\left(n + \frac{7}{4}\right) }. \end{align} When Stirling's approximation is used the expansion leads to $$ P_{n} = \Gamma\left(\frac{7}{4}\right) \, \frac{1}{n^{3/4}} \, \left(1 - \frac{21}{32} \, \frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right) \right). $$ From this result the limit can be stated as $$ \lim_{n \to \infty} \, P_{n} = \lim_{n \to \infty} \, \prod_{k=1}^{n} \left(\frac{4 \, k}{4 k + 3} \right) = 0. $$

Answer 4

There is a general method for this types of infinite products.See here or here.

Theorem: If $0<a_n<1$, $\prod_n(1-a_n)$ converts to a non zero number iff $\sum_na_n$ converges.

In your case, $\frac{4r}{4r+3}=1-\frac{3}{4r+3}$, and $\sum_r\frac{3}{4r+3}=\infty$; thus your infinite product converges improperly to $0$ (in the parlance of infinite products).