Convergence of $\sum_{t=1}^{\infty}{\left(\frac{1}{2}\right)^t} \ln\left(\frac{W+2^{t-1}-P}{W}\right)$ In a paper on the St Petersburg Paradox, it is said that the following sum converges: 
$\sum_{t=1}^{\infty}{\left(\frac{1}{2}\right)^t}  \ln\left(\frac{W+2^{t-1}-P}{W}\right)$
The author writes: "This sum converges (as long as each individual term is finite) as is readily shown using the ratio test." I've tried the ratio test, but I'm getting nowhere...
Thanks a lot for your help!
 A: Applying the Ratio Test,
$$\lim_{t\to\infty} \left|\frac{(1/2)^{t+1}\ln(\frac{W + 2^{t}-P}{W})}{(1/2)^{t}\ln(\frac{W + 2^{t-1}-P}{W})}\right| = \lim_{t\to\infty}\frac{1}{2} \left|\frac{\ln(\frac{W + 2^{t}-P}{W})}{\ln(\frac{W + 2^{t-1}-P}{W})}\right| = \lim_{t\to\infty}\frac{1}{2}\left|\frac{t\ln 2 - \ln W}{(t-1)\ln 2 - \ln W}\right| = \frac{1}{2} < 1$$
which shows the series converges.
To see why 
$$\lim_{t\to\infty}\frac{1}{2} \left|\frac{\ln(\frac{W + 2^{t}-P}{W})}{\ln(\frac{W + 2^{t-1}-P}{W})}\right| = \lim_{t\to\infty}\frac{1}{2}\left|\frac{t\ln 2 - \ln W}{(t-1)\ln 2 - \ln W}\right|$$
notice the following: $\ln x$ is a concave function, so $\ln(a + b) \leq \ln a + \ln b$. This means that $$\ln\left(\frac{2^t}{W}\right) < \ln\left(\frac{W + 2^t-P}{W}\right) < \ln\left(\frac{2^t}{W}\right) + \ln\left(\frac{W-P}{W}\right)$$
The constant term becomes insignificant as $t$ grows large; thus, $\ln\left(\frac{W + 2^t-P}{W}\right) \sim \ln\left(\frac{2^t}{W}\right)$. Because we are taking the limit as $t$ goes to $\infty$, I can drag the limit into the numerator and denominator and replace them with their asympotics.
A: \begin{align}
&\exp\left(\sum_{t=1}^{\infty}{\left(\frac{1}{2}\right)^t}  \ln\left(\frac{W+2^{t-1}-P}{W}\right)\right)
\\=&\prod_{t=1}^\infty \exp \left(2^{-t}  \ln\left(\frac{W+2^{t-1}-P}{W}\right)\right)
\\=&\prod_{t=1}^\infty \left(\frac{W+2^{t-1}-P}{W}\right)^{2^{-t}}.
\end{align}
This is bounded above by ${\prod_{t=1}^\infty \left(c2^{(t-1)}\right)^{}}^{2^{-t}}$ for some constant $c$, and
\begin{align}
\prod_{t=1}^\infty \left(c2^{(t-1)}\right)^{2^{-t}}=\prod_{t=1}^\infty c^{2^{-t}} 2^{(t-1)\times 2^{-t}}.
\end{align}
Clearly, $$\lim_{t\rightarrow \infty }c^{2^{-t}} 2^{(t-1)\times 2^{-t}}=1,$$
so the product converges, which implies that your sum converges.
