The limit of a logarithm of an infinite product series

Question

Please help me calculate this limit: $$ \lim_{n\rightarrow\infty}\ln\left(\prod_{k=1}^{n}\frac{2k+1}{2k}\right) $$

Answer 1

If we set: $$A_n=\prod_{k=1}^{n}\frac{(2k+1)}{2k}=\frac{(2n+1)!!}{(2n)!!}=\frac{(2n+1)!}{4^n n!^2}=(2n+1)\cdot\frac{1}{4^n}\binom{2n}{n}$$ we have: $$ A_n^2 = \prod_{k=1}^{n}\left(1+\frac{1}{k}+\frac{1}{4k^2}\right)\geq \prod_{k=1}^{n}\left(1+\frac{1}{k}\right) = n+1 $$ hence $$ A_n \geq \sqrt{n+1} $$ implies that our limit is $+\infty$, without invoking Stirling or Taylor.

Answer 2

It diverges, since $\sum_{k=1}^\infty \frac{1}{2k}$ diverges.

Without using theorems about infinite products, you can show pretty easily that $\log\left(1+\frac{1}{2k}\right) > \frac{1}{4k}$.

Specifically, since $\left(1+\frac{1}{2k}\right)^{4k}\to e^2$, for sufficiently large $k$, $$4k\log\left(1+\frac1{2k}\right)>1$$

More generally, if $x_i>0$, then $\prod_{i=1}^n (1+x_i) \geq 1+\sum_{k=1}^n x_i$. So if $\sum x_i$ diverges, then so does $\prod(1+x_i)$.

Answer 3

Hint: use $\ln(1+x)=x+O(x^2)$. So $\ln\frac{2k+1}{2k}=\ln(1+\frac1{2k})=\frac{1}{2k}+O(\frac{1}{k^2})$. What is $\sum_{k=1}\frac{1}{2k}$?

Answer 4

The partial product is

$$\frac{(2 n+1)!!}{2^n n!} = \frac{(2 n+1)!}{2^{2 n} n!^2}$$

which, in the limit, behaves as

$$\frac{(2 n+1)^{2 n+1} e^{-(2 n+1)} \sqrt{2 \pi (2 n+1)}}{2^{2 n} n^{2 n} e^{-2 n} 2 \pi n} \approx \frac{2 n (2 n)^{2 n} e e^{-(2 n+1)} \sqrt{2 \cdot 2 \pi n} }{2^{2 n} n^{2 n} e^{-2 n} 2 \pi n}$$

which simplifies to $2 \sqrt{\frac{n}{\pi}}$

As the log of this behaves as $\frac12 \log{n}$, the sequence diverges.

Answer 5

This limit goes to infinity. The product is a red herring, what you really have is $$ \lim_{n \to \infty} \sum_{k=1}^n \ln\left( 1 + \frac{1}{2k} \right)$$

We can expand the logs getting $$ \lim_{n \to \infty} \sum_{k=1}^n \left( \frac{1}{2k} - \frac{1}{8k^2} + O\left( \frac{1}{k^3}\right) \right) = \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{2k} - \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{8k^2} +\ldots $$ Except for the $\frac{1}{2k}$ term the others are all finite, so the sum (therefore the original product) diverges like $\sum_{k=1}^n \frac{1}{2k}$; the leading behavior is $\frac{1}{2}\ln n$.

Answer 6

I am surprised that nobody has applied the good-old-fashioned Gauss's test here. So, here we go. Write

$$\lim_{n\rightarrow\infty}\ln\left(\prod_{k=1}^{n}\frac{2k+1}{2k}\right)=\lim_{n\rightarrow\infty} \sum_{k=1}^{n} \ln\left(\frac{2k+1}{2k}\right)$$

and examine the terms of the series

$$a_k=\log\left(\frac{2k+1}{2k}\right)=\log\left(1+\frac{1}{2k}\right)$$

Now, the test ...

$$\frac{a_{k}}{a_{k+1}}= \frac{\log\left(1+\frac{1}{2k}\right)}{\log\left(1+\frac{1}{2k+2}\right)}=\frac{1+\frac{1}{2k}+O(k^{-2})}{1+\frac{1}{2k+2}+O(k^{-2})}=1+\frac{0}{k}+O(k^{-2})$$

Since the coefficient on $k^{-1}=0\le1$, the series diverges.