Asymptotic value of sum with complex numbers I have been looking for an asymptotic behavior for large $n$ of the following infinite sum of powers:
$$S_n = \displaystyle \lim_{N\rightarrow\infty} \sum_{k=1}^{N} \left[\left(1+i\,\frac{\alpha}{k}\right)^n-1\right]$$
where $\alpha = \log(2)/(2\pi) \simeq 1/10$.
I understand the imaginary number makes the sequence inside the series rotate, thus making it difficult to estimate the total sum. I am only interested in the real part. I don't need a very close approximation, but a loose bound better than $2^n$ would be ok.
Alternatively, I believe I could obtain a finite sum that yields the same result in terms of Bernoulli numbers $B_{2k}$:
$$S_n = -\frac{\pi}{2}\sum_{k=1}^{n/2}\left( \matrix{n \\ 2k} \right) \frac{B_{2k}\,\log^{2k}(2)}{(2k)!}$$
but that didn't take me too far either.
Thank you!
 A: Let us define the following for $\alpha \in (-1,1)$
$$S_n(\alpha) = \lim_{N \rightarrow \infty} \sum_{k=1}^N \left[\left(1 + \frac{i\alpha}{k}\right)^n - 1\right]$$
$$\implies \frac{\partial^j}{\partial\alpha^j}S_n(\alpha) = (i^j) (n)(n-1)\dots(n-j+1) \lim_{N \rightarrow \infty} \sum_{k=1}^N \left(1+\frac{i\alpha}{k}\right)^{n-j}\frac{1}{k^j}$$
For ease, let $S_n^j(\alpha) = \frac{\partial^j}{\partial\alpha^j}S_n(\alpha)$
$$\implies S_n^j(0) = (i^j)(n)(n-1)\dots(n-j+1)\zeta(j)$$
To make the calculations easier, let us do calculation for $2n$ instead.
\begin{align}
S_{2n}(\alpha) &= \frac{S_{2n}(0)}{0!} + \frac{S_{2n}^1(0)\alpha}{1!} + \frac{S_{2n}^1(0)\alpha^2}{2!} + \dots\\
&= \sum_{j=1}^{2n} (i^j)(2n)(2n-1)\dots(2n-j+1)\zeta(j)\frac{\alpha^j}{j!}\\
&= \sum_{j=1}^{2n} (i^j)\binom{2n}{j}\zeta(j)\alpha^j\\
\operatorname{Re}(S_{2n}(\alpha)) &= \sum_{j=1}^{n} (-1)^j \binom{2n}{2j}\zeta(2j)\alpha^{2j}
\end{align}
We also have,
$$2\le\zeta(2j) + \zeta(2j+2)\le 2\zeta(2j) \le \pi^2/3 \,\,\, \forall j\ge1$$
So for large values of $n$,
$$\implies \frac{1}{2}\left((1+i\alpha)^{2n} + (1-i\alpha)^{2n} - 2\right)\le \operatorname{Re}(S_{2n}(\alpha)) \le \frac{\pi^2}{6}\left((1+i\alpha)^{2n} + (1-i\alpha)^{2n} - 2\right)$$
This is only slightly better than the $2^n$ bound but hope this helps.
A: $$S_n = \sum_{k=1}^{N} \left[\left(1+i\,\frac{\alpha}{k}\right)^n-1\right]$$ Using the asymptotics of generalized harmonic numbers, we have for large values of $N$
$$S_n= n i \alpha (\log(N)-  \gamma)-\frac {n(n-1)}{12}\pi^2 \alpha^2+\cdots+O\left(\frac{1}{N}\right)$$
