Alternating sum of logarithms

Question

I'm trying to solve the product $\prod_{k=1}^{\infty} \frac{2k}{2k+1}$:
I've started by letting it be equal to P:
P = $\prod_{k=1}^{\infty} \frac{2k}{2k+1}$ $\implies$ ln(P) = $\sum_{k=1}^{\infty} \ln(\frac{2k}{2k+1})$
So I'm interested in the above sum of logarithms. Having manipulated the summand a little:
$\sum_{k=1}^{\infty} \ln(\frac{2k}{2k+1})$ = $\sum_{k=1}^{\infty} (\ln(2k) - \ln(2k+1))$

Expanding the sum out:
$\sum_{k=1}^{\infty} (\ln(2k) - \ln(2k+1))$ = ln(2) - ln(3) + ln(4) - ln(5) + ln(6) ... = $\sum_{k=2}^{\infty} (-1)^k \ln(k)$
And this is where I'm stuck. I've attempted substituting the integral definition of the logarithm like so:
$\sum_{k=2}^{\infty} (-1)^k \ln(k)$ = $\sum_{k=2}^{\infty} \int_{1}^{k} \frac{1}{t}\ dt$
This hasn't led me anywhere. What should I do to solve this sum? I suspect a Laurent Series, but I haven't studied them enough to really know how to even begin implementing one to solve the problem.

Answer 1

Cross multiplication shows that $$ \left(\frac{2k}{2k+1}\right)^2\le\frac{k}{k+1}\tag1 $$ Thus $$ \begin{align} \prod_{k=1}^n\left(\frac{2k}{2k+1}\right)^2 &\le\prod_{k=1}^n\frac{k}{k+1}\tag{2a}\\ &=\frac1{n+1}\tag{2b} \end{align} $$ Therefore, $$ \prod_{k=1}^n\frac{2k}{2k+1}\le\frac1{\sqrt{n+1}}\tag3 $$