Compute $\lim_{n \rightarrow \infty} \left(1 + \frac{i}{n^2 + in}\right)^n$

Question

I have a feeling that

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{i}{n^2 + in}\right)^n = 1,$$

but I don't know how to justify it. If $i$ was a real number $r > 0$, I know how to compute this: take logs and then solve $$\lim_{n \rightarrow \infty} n \ln \left( 1 + \frac{r}{n^2 + rn}\right)$$ which after applying L'Hopital's rule gives $0$, and hence an original limit of $e^0 = 1$. But since complex numbers are involved I'm don't even know if I'm allowed to apply the same logic.

What theorems / results can be applied to efficiently solve the limit, now that complex numbers are involved?

Answer 1

More generally, if $\lim_{n\to \infty} ng(n)=0$ then $$\lim_{n\to\infty} (1+g(n))^n = 1.$$

This follows from a simple application of the binomial theorem.

Proof: When $|ng(n)|<1$ you have:

$$\begin{align}\left|(1+g(n))^n - 1\right| &\leq \sum_{k=1}^n \binom{n}{k}\left|g(n)\right|^k\\ &\leq \sum_{k=1}^n n^k|g(n)|^k \\&= \sum_{k=1}^\infty (n|g(n)|)^k\\&=\frac{n|g(n)|}{1-|ng(n)|} \end{align}$$

So $(1+g(n))^n\to 1$ since $\frac{ng(n)}{1-ng(n)}\to 0$.

See this answer for more about applying this result.

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In your case, $g(n)=\frac{i}{n^2+in}$ then $ng(n)=\frac{i}{n+i}\to 0.$

Answer 2

It works almost like the case when $i$ is replaced by a real number. You can check that the real part of $1+\frac i {n^{2}+in}$ is positive. Hence, its principal logarithm $Log (1+\frac i {n^{2}+in})$ is well defined. Also, $Log (1+z)\sim z$ as $z \to 0$. Hence, $nLog (1+\frac i {n^{2}+in}) \to 0$ and the given limit is $e^{0}=1$.