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
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0Are you willing to use L'Hospital's Rule? That will prove the limit is $2\pi i$ and you will not need to directly prove anything involving norms. Rather, those things will be a consequence of having established the limit. – 2012-02-02
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0Isn't it true that sequences don't always converge to their limit? I can find the limit but I feel like I need to prove the sequence converges to that limit. – 2012-02-02
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0@Ashley: By definition, a sequence converges if and only if it has a limit. To say that $(x_n)$ converges means that there exists $x\in\mathbb C$ such that $\lim\limits_{n\to\infty} x_n=x$. "The limit of the sequence $(x_n)$ is $x$," and "The sequence $(x_n)$ converges to $x$," mean precisely the same thing. – 2012-02-02
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0If a sequence converges at all, then the definition of its _limit_ is the value to which it converges. And if the sequence does not converge, then the limit does not exist. (There are also sequences that are called _Cauchy_, where the terms get arbitrarily close to each other, but not necessarily to any one specific value.) The good news is that you might not have a math problem, so much as a vocabulary misunderstanding. – 2012-02-02
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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