The limit of a logarithm of an infinite product series Please help me calculate this limit: 
$$ \lim_{n\rightarrow\infty}\ln\left(\prod_{k=1}^{n}\frac{2k+1}{2k}\right) $$
 A: 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 $\prod(1+x_i)$.
A: 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.
A: 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.
A: 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.
A: 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 $\sum_{k=1}\frac{1}{2k}$?
A: 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 $\frac{1}{2}\ln n$.
