TITLE: Proofs of $\lim\limits_{n \to \infty} \left(H_n - 2^{-n} \sum\limits_{k=1}^n \binom{n}{k} H_k\right) = \log 2$
QUESTION [32 upvotes]: Let $H_n$ denote the $n$th harmonic number; i.e., $H_n = \sum\limits_{i=1}^n \frac{1}{i}$.  I've got a couple of proofs of the following limiting expression, which I don't think is that well-known: $$\lim_{n \to \infty} \left(H_n - \frac{1}{2^n} \sum_{k=1}^n \binom{n}{k} H_k \right) = \log 2.$$
I'm curious about other ways to prove this expression, and so I thought I would ask here to see if anybody knows any or can think of any.  I would particularly like to see a combinatorial proof, but that might be difficult given that we're taking a limit and we have a transcendental number on one side.  I'd like to see any proofs, though.  I'll hold off from posting my own for a day or two to give others a chance to respond first.
(The probability tag is included because the expression whose limit is being taken can also be interpreted probabilistically.)

(Added: I've accepted Srivatsan's first answer, and I've posted my two proofs for those who are interested in seeing them.  
Also, the sort of inverse question may be of interest.  Suppose we have a function $f(n)$ such that  $$\lim_{n \to \infty} \left(f(n) - \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} f(k) \right) = L,$$ where $L$ is finite and nonzero.  What can we say about $f(n)$?  This question was asked and answered a while back; it turns out that $f(n)$ must be $\Theta (\log n)$.  More specifically, we must have $\frac{f(n)}{\log_2 n} \to L$ as $n \to \infty$.)

REPLY [7 votes]: Here are my two proofs, in case anyone is interested.

Proof 1: The $n$th term is $\sum_{k=1}^n \frac{1}{k 2^k}$.
Let $\Delta f(k) = f(k+1) - f(k)$, i.e., the finite difference of $f(k)$.
Let $B(n) = \sum_{k=0}^n \binom{n}{k} f(k)$, $A(n) = \sum_{k=0}^n \binom{n}{k} \Delta f(k)$.
A few years ago I proved$^1$ the following relationship between $B(n)$ and $A(n)$:
$$B(n) = 2^n \left(f(0) + \sum_{k=1}^n \frac{A(k-1)}{2^k}\right).$$
Now, $\Delta H_n = \frac{1}{n+1}$.  And we know that 
$$\sum_{k=0}^n \binom{n}{k} \frac{1}{k+1} = \frac{1}{n+1} \sum_{k=0}^n \binom{n+1}{k+1} = \frac{2^{n+1}-1}{n+1},$$ 
using the fact that $\binom{n}{k} \frac{n+1}{k+1} = \binom{n+1}{k+1}$.
Thus the formula above for $B(n)$ yields 
$$\sum_{k=0}^n \binom{n}{k} H_k = 2^n \sum_{k=1}^n \frac{2^k-1}{k2^k} = 2^n\left(H_n - \sum_{k=1}^n \frac{1}{k2^k}\right).$$
Rearranging, we get 
$$H_n - \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} H_k = \sum_{k=1}^n \frac{1}{k 2^k}.$$
Thus $$\lim_{n \to \infty} \left(H_n - \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} H_k\right) = \sum_{k=1}^{\infty} \frac{1}{k 2^k} = \log 2,$$
where the $\log 2$ is obtained, as in Srivatsan's second answer, by substituting $-1/2$ into the Maclaurin series for $\log(1+x)$.
(This is how I first came across the limit result; I was trying to find an expression for $\sum_{k=0}^n \binom{n}{k} H_k$.)
$^1$"Combinatorial Sums and Finite Differences," Discrete Mathematics, 307 (24): 3130-3146, 2007. 

Proof 2: The probabilistic argument.
Let $X_1, X_2, \ldots, X_n$ be $n$ independent and identically distributed exponential random variables with rate parameter $\lambda$.  The minimum of these $n$ random variables is known to have an exponential $n \lambda$ distribution.  Let $X_{(k)}$ denote the $k$th smallest.  Because of the memoryless property of the exponential distribution, $X_{(k+1)} - X_{(k)}$ has an exponential $(n-k) \lambda$ distribution.  So $E[X_{(k+1)} - X_{(k)}] = \frac{1}{(n-k)\lambda}$.  Let $Y_n = X_{(n)}$ (i.e., the largest).  Thus $$E[Y_n] = \sum_{k=0}^{n-1} E[(X_{(k+1)} - X_{(k)})= \sum_{k=1}^n \frac{1}{\lambda k} = \frac{H_n}{\lambda}.$$ 
Now, consider $E[Y_n | Y_n > 1]$.  Since $P(X_i < 1) = 1-e^{-\lambda}$, conditioning on the number of the $X_i$ with values less than $1$ yields $$E[Y_n | Y_n > 1] = 1 + \sum_{k=0}^{n-1} \binom{n}{k} \left(1 - e^{-\lambda}\right)^k \left(e^{-\lambda}\right)^{n-k} \frac{H_{n-k}}{\lambda}$$
$$= 1 + \sum_{k=1}^n \binom{n}{k} \left(1 - e^{-\lambda}\right)^{n-k} \left(e^{-\lambda}\right)^k \frac{H_k}{\lambda}.$$
But as $n \to \infty$, $E[Y_n] - E[Y_n|Y_n > 1] \to 0$, since $P(Y_n > 1) = 1 - e^{-\lambda n}$.  Therefore,
$$\lim_{n \to \infty} \left(\frac{H_n}{\lambda} - \sum_{k=1}^n \binom{n}{k} \left(1 - e^{-\lambda}\right)^{n-k} \left(e^{-\lambda}\right)^k \frac{H_k}{\lambda}\right) = 1.$$
Letting $\lambda = \log 2$ yields the result.