Value of $\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$ Is there a closed form expression of
$$\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=1}^{2^n} \{\log_{2}k\}$$
, where $\{x\}=x-\lfloor x \rfloor$ denotes the fractional part of $x$? $\text{I think this limit will converge to a point between $0$ and $\log_{2}3-1$}$
   by $(x+y)(x-y)=(x^2-y^2)$,  but I can't go further.
 A: By periodicity, $\log_2 (k)$ and $\log_2 (k) - n$ has the same fractional part. Hence
\begin{align*}
\lim_n \frac 1{2^n} \sum_1^{2^n} \{\log_2(k)\} &= \lim_n \frac 1{2^n} \sum_1^{2^n} \left\{\log_2 \left(\frac k{2^n}\right) \right\}  \\&= \lim_n \frac 1{2^n} \sum_1^{2^n} \log_2 \left( \frac k{2^n }\right) - \left\lfloor \log_2\left(\frac k{2^n}\right)\right\rfloor \\
&= \int_0^1 \log_2(x) \mathrm dx - \lim f(n)/2^n,
\end{align*}
where
$$
f(n) = \sum_1^{2^n} \lfloor \log_2(k/2^n)\rfloor = \sum_1^{2^n} \lfloor \log_2(k)\rfloor - 2^n \cdot n.
$$
Now calculate the sum of the integer parts. Since $\lfloor \log_2(k)\rfloor = j$ when $2^j \leq k \leq 2^{j+1}-1$, we have
$$
f(n) + 2^n \cdot n = 0 + n + \sum_0^{n-1} 2^j j.
$$
Let $S(n) = \sum_0^{n-1}j 2^j$, then $2S(n) = \sum_0^{n-1} 2^{j+1} j = \sum_1^n 2^j (j-1)$, thus
\begin{align*}
S(n) = 2S(n) - S(n) &= \sum_1^n 2^j (j-1) - \sum_0^{n-1} 2^j j= \sum_2^n 2^j (j-1) - \sum_1^{n-1} 2^j j \\
&= 2^n (n-1) + \sum_2^{n-1} 2^j (-1) -2 = 2^n (n-1) -2(2^{n-1}-1) \\
&= 2^n(n-2) +2,
\end{align*}
then 
$$
\frac {f(n)}{2^n} = \frac {2^n (n-2) + 2 - 2^n n-n}{2^n} = -2 +\frac {2-n}{2^n} \to -2. 
$$
Hence the result is
$$
\int_0^1 \log_2 (x)\mathrm d x + 2 = 2 - \frac 1 {\log(2)}. 
$$
