I have taken the limit of $x_n$ and got $2i\pi$. Now I am stuck trying to show $|n(e^{\frac{2\pi i}{n}}-1)-2i\pi|=0$. I am thinking I should try to write this in the form of $|a+bi|$ but I can't figure out how.
How do you show $x_n=n(e^{\frac{2\pi i}{n}}-1)$ converges or not in C with the usual norm?
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sequences-and-series
complex-numbers
convergence-divergence
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1I see where you are coming from though, given how questions can be worded. Like, "Let $L=2$. Prove that $\frac{2n+1}{n+1}$ converges to $L$." It seems to imply that we already know the limit is $2$, and you have to prove a "separate" convergence statement. Really, the problem is calling a number by $L$ with foresight that we _will later_ know that $2$ is the $L$imit. Anyway, with your current problem all of the work for the convergence proof has already gone into the proof of L'Hospital's Theorem, and the proof that if a real function converges, then so do sequences that use that function. – 2012-02-02
2 Answers
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Let $f(z)=e^{2\pi i z}$. Then $\displaystyle{x_n=\frac{f\left(\frac{1}{n}\right)-f(0)}{\frac{1}{n}}}$. You can therefore find $\lim\limits_{n\to\infty}\, x_n$ by finding a derivative.
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How about using that $ n(e^{2\pi i /n} -1) = \frac{e^{2\pi i /n} -1}{1/n - 0}$ which looks a lot like a differential quotient. Hence letting $1/n \to 0$ (i.e. $ n\to \infty$) you should get $ n(e^{2\pi i /n} -1) \to 2\pi i $