Convergence of $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\binom{n}{k}\frac{\sin k}{k}$ Is this series convergent? I met it when I was studying some fractals.
$$\lim_{n\to\infty}\sum_{k=1}^{n}\binom{n}{k}\frac{\sin k}{k}$$  
 A: A very long comment that is made to make you lose faith in the convergence of your limit ; I didn't actually prove the fact that it diverges though. For $x \in \mathbb C \backslash \{0\}$, we have
$$
\begin{align}
\sum_{k=1}^n \binom nk \frac{x^k}k &= \sum_{k=1}^n \binom nk\int_0^x t^{k-1} \, \mathrm dt \\
 &= \int_0^x \left( \sum_{k=1}^n \binom nk t^{k-1}  \right) \, \mathrm dt \\
 &= \int_0^x \left( \frac 1t \left[\sum_{k=0}^n \binom nk t^k - 1 \right] \right) \mathrm dt \\
 &= \int_0^x \frac{(1+t)^n-1}{t} \, \mathrm dt \\
 &= \int_0^x \frac{(1+t)^n - 1}{(1+t)-1} \, \mathrm dt \\
 &= \int_0^x \sum_{k=0}^{n-1} (1+t)^k \, \mathrm dt \\
 &= \sum_{k=0}^{n-1} \int_0^x (1+t)^k \, \mathrm dt \\
 &= \sum_{k=0}^{n-1} \left[ \frac{(1+x)^{k+1} - 1}{k+1} \right] \\
 &= \sum_{k=1}^n \left[ \frac{(1+x)^k - 1}{k} \right].
\end{align}
$$
hence letting $x = e^i$, we get
$$
\sum_{k=1}^n \left[ \frac{(1+e^i)^k - 1}{k} \right] = \sum_{k=1}^n \binom nk \frac{e^{ik}}k = \left( \sum_{k=1}^n \binom nk \frac{\cos k}{k} \right) + i \left( \sum_{k=1}^n \binom nk \frac{\sin k}{k} \right) 
$$
If the series for $\sin$ and $\cos$ would converge, so would the series 
$$
\sum_{k \ge 1} \frac{(1+e^i)^k - 1}{k}.
$$
The main term of this series doesn't go to zero because
$$
\left| \frac{(1+e^i)^k - 1}{k} \right| \ge \frac{|1+e^i|^k - 1}{k} \ge \frac{(1.75)^k - 1}{k} \to \infty.
$$
(I used the fact that $|1+e^i| \approx 1.755165...$ see here.) So we know that either the real part or the imaginary part diverges. I think it's enough to lose faith in having either one converge since they behave similarly (and the series considered isn't just divergent, its main term goes to infinity...)
Hope that helps,
A: Before taking the limit, the sum has an expression form in terms of hypergeometric functions:
$$
\sum _{k=1}^n \binom{n}{k}\frac{\sin k }{k} = 
-\frac{1}{2} i e^{-i} n \left(e^{2 i} \, _3F_2\left(1,1,1-n;2,2;-e^i\right)-\, _3F_2\left(1,1,1-n;2,2;-e^{-i}\right)\right)
$$
Now you just need to study the asymptotics of such functions, namely, do they decay fast enough as the thrid argument goes to $-\infty$.  I think you could find these properties in the usual references about special functions (DLMF or Abramowitz and Stegun), and that would suffice as a proof.  I can certainly say that they grow unboundedly as the third argument becomes increasingly negative, but you'd need some asymptotic bounds to finish off a proof.
