The sequence z_n=dfrac1n+e^(nπ i/3) has six accumulation…
The sequence $z_n=\dfrac1n+e^{n\pi i/3}$ has six accumulation points. Determine them.
Explanation of Method
The term $\frac1n\to0$ contributes nothing in the limit, while $e^{n\pi i/3}$ is periodic in $n$ with period $6$ (since $\frac{2\pi}{\pi/3}=6$); split the sequence into six subsequences by $n\bmod6$ and take the limit of each.
Step 1 · Period of $e^{n\pi i/3}$
The values for $n=1,\ldots,6$ are the six distinct sixth roots of unity, then the cycle repeats: $e^{(n+6)\pi i/3}=e^{n\pi i/3}e^{2\pi i}=e^{n\pi i/3}$.
Step 2 · Limit along each residue class mod 6
For each fixed residue $r\in\{1,2,3,4,5,0\}$, the subsequence $n\equiv r\pmod6$ has:
$$ z_n=\frac1n+e^{r\pi i/3}\ \xrightarrow[n\to\infty]{}\ 0+e^{r\pi i/3}=e^{r\pi i/3} $$
since $\frac1n\to0$ along any such subsequence while the exponential term stays fixed at $e^{r\pi i/3}$.
Step 3 · List all six limits
$$ \begin{array}{ll} r=1: & e^{\pi i/3}=\frac12+i\frac{\sqrt3}2 \\ r=2: & e^{2\pi i/3}=-\frac12+i\frac{\sqrt3}2 \\ r=3: & e^{\pi i}=-1 \\ r=4: & e^{4\pi i/3}=-\frac12-i\frac{\sqrt3}2 \\ r=5: & e^{5\pi i/3}=\frac12-i\frac{\sqrt3}2 \\ r=0: & e^{2\pi i}=1 \end{array} $$
The six accumulation points form a regular hexagon inscribed in the unit circle.
Verification
These are exactly the six solutions of $w^6=1$: substituting $r=0,\ldots,5$ into $e^{r\pi i/3}$ sweeps out all sixth roots of unity, each spaced $60^\circ$ apart, consistent with $\left(e^{r\pi i/3}\right)^6=e^{2r\pi i}=1$.
Result
The vanishing term $\frac1n$ only nudges each subsequence toward its limiting sixth root of unity without changing which six points are approached — exactly the six vertices of a regular hexagon on the unit circle.