# How to evaluate $\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n$ where $i=\sqrt{-1}$?

How to evaluate the following limit? $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n$$ Here $$i=\sqrt{-1}$$.

I got: $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n = \lim_{n\to \infty} \frac{(i-1)^n}{n(\sqrt{2})^n}$$ I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?

• It's very hard to understand what you wrote. Try mathjax – DonAntonio Nov 24 '18 at 21:38
• Express $(1+i)/\sqrt{2}=\exp(i\pi/4)$. – Diger Nov 24 '18 at 21:39
• L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $\sqrt{blah blah}$. A limit of a function f(x) from $n \to \infty$ can be written $\lim_{n\to \infty} f(x)$. Just a few pointers. – Namaste Nov 24 '18 at 21:43
• If your function $f(x)$ is a fraction, you can write it as follows $f(x) = \frac{"numerator here"}{"denominator here"}$ – Namaste Nov 24 '18 at 21:46
• Write $\frac{1+i}{\sqrt{2}}=e^{i\pi/4}\implies (\frac{1+i}{\sqrt{2}})^n=e^{i\ n\pi/4}$ – Shubham Johri Nov 24 '18 at 21:57

## 2 Answers

We have $$\left|\frac{1+i}{\sqrt{2}}\right|=1$$ Thus $$\left|\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^{\!n}\right|=\frac{1}{n}$$

• egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2 – KarmaL Nov 25 '18 at 9:25
• @L.Spy $\sqrt{(1/\sqrt{2})^2+(1/\sqrt{2})^2}=\sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$. – egreg Nov 25 '18 at 9:40

$$\lim_{n\to \infty} |\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n|=\lim_{n\to \infty} {1\over n}=0$$therefore $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n=0$$