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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!

Sohom Paul
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Julian Rodemann
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2 Answers2

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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.

Sohom Paul
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  • Thank you!! Could you please specify how to get from $$ lim_{t\to\infty}\frac{1}{2} \left|\frac{\ln(\frac{W + 2^{t}-P}{W})}{\ln(\frac{W + 2^{t-1}-P}{W})}\right| $$ to $$ \lim_{t\to\infty}\frac{1}{2}\left|\frac{t\ln 2 - \ln W}{(t-1)\ln 2 - \ln W}\right|? $$ I don't understand why we can ignore W and P in the numerator. – Julian Rodemann Jul 04 '19 at 15:37
  • The idea is to notice that $\ln \left(\frac{W + 2^t - P}{W}\right) \sim \ln \left(\frac{2^t}{W}\right)$, so in the limit, we can throw away these terms. If you like me to show more work or be more rigorous, please let me know! – Sohom Paul Jul 04 '19 at 15:47
  • Or perhaps more directly, the logarithm is a concave function. I'll edit my post for these extra steps. – Sohom Paul Jul 04 '19 at 15:52
  • Thank you very, very much! I got it now. – Julian Rodemann Jul 04 '19 at 19:20
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\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.

cangrejo
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  • Thanks a lot! One short question: Why is bounded above by ${\prod_{t=1}^\infty \left(c2^{(t-1)}\right)^{}}^{2^{-t}}$? Isn't it rather bounded above by ${\prod_{t=1}^\infty \left(c + 2^{(t-1)}\right)^{}}^{2^{-t}}$ ? – Julian Rodemann Jul 04 '19 at 15:41
  • Note that you can express any sum $2+a$ as a factor of $2$. The bound I used grows with $t$, so it's not very tight, but it still holds and it's very convenient in this case, because the exponent is distributive with respect to the factors. This makes it easy to analyze. – cangrejo Jul 04 '19 at 18:06