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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?

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    $\begingroup$ It's very hard to understand what you wrote. Try mathjax $\endgroup$ – DonAntonio Nov 24 '18 at 21:38
  • $\begingroup$ Express $(1+i)/\sqrt{2}=\exp(i\pi/4)$. $\endgroup$ – Diger Nov 24 '18 at 21:39
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    $\begingroup$ 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. $\endgroup$ – amWhy Nov 24 '18 at 21:43
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    $\begingroup$ If your function $f(x)$ is a fraction, you can write it as follows $f(x) = \frac{"numerator here"}{"denominator here"}$ $\endgroup$ – amWhy Nov 24 '18 at 21:46
  • $\begingroup$ Write $\frac{1+i}{\sqrt{2}}=e^{i\pi/4}\implies (\frac{1+i}{\sqrt{2}})^n=e^{i\ n\pi/4}$ $\endgroup$ – Shubham Johri Nov 24 '18 at 21:57
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We have $$ \left|\frac{1+i}{\sqrt{2}}\right|=1 $$ Thus $$ \left|\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^{\!n}\right|=\frac{1}{n} $$

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  • $\begingroup$ egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2 $\endgroup$ – KarmaL Nov 25 '18 at 9:25
  • $\begingroup$ @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 $1$. $\endgroup$ – egreg Nov 25 '18 at 9:40
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$$\lim_{n\to \infty} |\frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n|=\lim_{n\to \infty} {1\over n}=0$$therefore $$\lim_{n\to \infty} \frac{i}{n}\left(\frac{1+i}{\sqrt{2}}\right)^n=0$$

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