Interesting identity, deduced from a Probability problem of Ming (CU,CSC)

I can do all except (b) orz.

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Well this is a good example that why combinatorics and probability can generate interesting identity. (discussion from part (e) on) It can be easily deduced that $\displaystyle P(B_i)=\frac{1}{2^{n-1}(2^n-1)}$. Now we turn our objective to finding the probability that $A$ and $B$ will meet at the $r$-th round. Having known what $P(B_i)$ is, our desired probability is obviously

$p = \displaystyle 2^{n-r}\cdot P(B_i) = \frac{2^{n-r}}{2^{n-1}(2^n-1)}$.

We take a step further, the same probability can be duduced if $A$ and $B$ cannot meet and they win at the first $r-1$-th rounds and occasionally meet at the $r$-th round, this is the same as $$p = \underbrace{\prod_{k=1}^{r-1}\left\{\frac{\displaystyle \binom{2^{n-k}}{2}\cdot 2^3}{2^{n-k+1}(2^{n-k+1}-1)}\cdot \left(\frac{1}{2}\right)^2\right\}}_{\text{don't meet and win at the first r - 1 rounds}}\cdot \overbrace{\left(\frac{2}{2^{n-r+1}}\cdot \frac{1}{2^{n-r+1}-1}\cdot 2^{n-r}\right)}^{\text{meet at the last round}}.$$ We now equate two $p$'s, having $$\prod_{k=1}^{r-1}\left(1-\frac{1}{2^{n-k+1}-1}\right)=\frac{2^n-2^{r-1}}{2^n-1},$$ by simple algebra we get a more convenient form $$\boxed{\dis \prod_{k=j}^n\left(1-\frac{1}{2^k-1}\right)=1-\frac{2^{n-j+1}-1}{2^n-1}.}$$ If this is an unprecedented identity, then I would call it CC's identity (政昌恆等式).

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Motivated by this problem, if we don't know the above formula, can we show that $$\sum\limits_{k=2}^\infty\displaystyle\ln\left(1-\frac{1}{2^k-1}\right)$$ converges? The answer is yes! Take a suitable $N$ such that when $k >N$, we always have $\displaystyle \frac{1}{2^k - 1}<1$, consider the taylor expansion $\displaystyle \ln (1+x)=x-\frac{1}{(1+c)^2}x^2$, for some $c$ in between $0$ and $x$, now consider $|x|<1$ such that $|c|<\delta <1$, then we get by putting $x=x_k$, $|x_k|<1$, $$\frac{x_k^2}{(1+\delta)^2}<x_k-\ln(1+x_k)<\frac{x_k^2}{(1-\delta)^2},$$ actually $x>\ln (1+x),\forall x>-1$ simply because $e^x\ge 1+x$. Now let $\displaystyle x_k=-\frac{1}{2^k-1}$, our required $N$ is simply 1, so we know that $\sum_{k=2}^\infty x_k^2$ converges, thus by above inequality, $\sum_{k=2}^\infty (x_k-\ln(1+x_k))$ also converges by comparison test, since $\sum_{k=2}^\infty x_k$ converges, $\sum_{k=2}^\infty \ln(1+x_k)$ also converges, we are done. Moreover, $\sum\limits_{k=2}^\infty\displaystyle\ln\left(1-\frac{1}{2^k-1}\right)=-\ln 2$.