Find limlimits_(n→∞)n(dfrac1+i2)^n
Find $\ \lim\limits_{n\to\infty}n\left(\dfrac{1+i}2\right)^n$.
Explanation of Method
Bound the modulus $|r|<1$, then apply the general fact that exponential decay always beats polynomial growth: $n^k|r|^n\to0$ for any fixed $k$ whenever $|r|<1$.
Step 1 · Modulus of the base
$$ r=\frac{1+i}2, \qquad |r|=\frac{\sqrt2}2=\frac1{\sqrt2}\approx0.707<1 $$
Step 2 · Reduce to a real limit
$$ \left|n\left(\frac{1+i}2\right)^n\right|=n|r|^n=\frac n{2^{n/2}} $$
Step 3 · Show $\dfrac n{2^{n/2}}\to0$ (ratio test)
Let $a_n=n\left(\frac1{\sqrt2}\right)^n$. Then:
$$ \frac{a_{n+1}}{a_n}=\frac{n+1}n\cdot\frac1{\sqrt2}\ \xrightarrow[n\to\infty]{}\ \frac1{\sqrt2}<1 $$
Since the ratio tends to a value strictly below $1$, $a_n\to0$.
Step 4 · Conclude
Since $\left|n\left(\frac{1+i}2\right)^n\right|=a_n\to0$:
$$ \boxed{\lim_{n\to\infty}n\left(\frac{1+i}2\right)^n=0} $$
General principle
For any complex $z$ with $|z|<1$ and any polynomial $p(n)$ (of any fixed degree): $\ p(n)z^n\to0$, because a fixed exponential decay rate always eventually dominates any fixed polynomial growth rate.
Result
Since $\left|\frac{1+i}2\right|=\frac1{\sqrt2}<1$, the geometric decay overwhelms the linear factor $n$, driving the whole product to zero.