Prove that $\lim_{n \to \infty}n\left(\frac{1+i}{2}\right)^n = 0$ I understand for this proof I must use the principle that I must find $N$, such that for $|u_n - 0| < \epsilon$ I have $n > N$ for $N$ that depends on $\epsilon$. However, I can't seem to get $n$ alone.
$$\left|n\left(\frac{1+i}{2}\right)^n\right| < \epsilon$$ led me to $$\left|\ln(n) + n(\ln(1+i) - \ln(2))\right| < \ln(\epsilon).$$ I don't know how to take this farther. 
 A: Hint: Use (or first prove) that if $\,a_n\ge 0\;$ and decreasing monotonically to zero and $\;\sum a_n\;$ converges then
$$\lim_{n\to\infty} na_n=0$$
A: Prove the modulus $\left|n\left(\frac{1+i}{2}\right)^n\right|\rightarrow0$.
$$\begin{align}
\left|n\left(\frac{1+i}{2}\right)^n\right| <& \epsilon\\
n \left(\frac{1}{\sqrt2}\right)^n <& \epsilon\\
n e^{-.5n\ln2} <& \epsilon\\
-.5n(\ln2) e^{-.5n\ln2} >& -.5\epsilon\ln2\\
\end{align}$$
For small enough $\epsilon$ ($-.5\epsilon\ln2 \ge -\frac{1}{e}$, or $\epsilon\le\frac{2}{e\ln2}\approx1.06$) and large enough $n$ ($-.5n\ln2 \le -1$, or $n\ge2\ln2>1$), the left hand side can be written into $f(w)=we^w$ where $w < -1$ and $f(w) \ge -\frac{1}{e}$. There is an inverse of this function know as the lower branch of Lambert W function $W_{-1}(x)$. The inverse function is decreasing for the domain $\left[-\frac{1}{e},0\right)$. Hence
$$\begin{align}
-.5n(\ln2) e^{-.5n\ln2} >& -.5\epsilon\ln2\\
-.5n\ln2 <& W_{-1}(-.5\epsilon\ln2)\\
n >&-\frac{2 W_{-1}(-.5\epsilon\ln2)}{\ln2}
\end{align}$$
For larger $\epsilon$ that makes $-.5\epsilon\ln2 < -\frac{1}{e}$, the inequality is always true and one can simply take $n > 0$.
