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Is it true that $\sum\limits_{n=1}^N e^{in} = \frac{e^i(1-e^{iN})}{1-e^i}$ even if $| e^i | > 1$?

I know this question is quite trivial and I will understand if it gets removed.

I am trying to show that $\sum\limits_{n=1}^\infty \frac{cos(n)}{\sqrt{n}}$ is convergent by using the Dirichlet's test. And taking, for the one series, $cos(n)$, now

$|\sum\limits_{n=1}^N cos(n) |< |\sum\limits_{n=1}^N (cos(n) + i sin(n))| = |\sum\limits_{n=1}^N e^{in}|$

And I am trying to find a value for $|\sum\limits_{n=1}^N e^{in}|$

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    $\begingroup$ That is an identity and it holds no matter what integer $N$ you put in. However, the identity in the limit $N\to \infty$ is another thing. $\endgroup$ – Winther Oct 17 '14 at 0:46
  • $\begingroup$ I am not taking $N \rightarrow \infty$, I just need to show that $|\sum\limits_{n=1}^N cos(n)|$ is bounded above for the Dirichlet test to work. $\endgroup$ – Pablo Oct 17 '14 at 0:52
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Yes, the statement $$ \sum\limits_{n=1}^N e^{in} = \frac{e^i(1-e^{iN})}{1-e^i} $$ for any choice of $e \in \Bbb C$. The only issue you have is whether the limit exists when $N \to \infty$.

Note that the proof (of this and the general geometric series formula) is not particularly difficult. We have $$ \sum\limits_{n=1}^N e^{in} = e^{i} + e^{2i} + \cdots + e^{iN}\\ e^i\sum\limits_{n=1}^N e^{in} = e^{2i} + \cdots + e^{iN} + e^{i(N+1)} $$ Subtract the two, factor, and solve.

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$|e^{it}|$ is always equal to $1$, for any $t$. To see this note that $e^{it}\overline{e^{it}}=e^{it}e^{-it}=1$. Now, if a complex number $z$ satisfies $z\overline{z}=1$, then by definition, $|z|^2=1$ and hence $|z|=1$.

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  • $\begingroup$ Alright, but then can I use the geometric series with $\sum\limits_{n=1}^N e^{in}$? I think I would rather try something along the lines of $\sum\limits_{n=1}^N|e^{in}| = \sum\limits_{n=1}^N 1 = N$, do you think this could work? I mean, N is a finite constant so I think with the Dirichlet test it could work. $\endgroup$ – Pablo Oct 17 '14 at 0:48
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    $\begingroup$ @Pablo: Your approach looks right to me at least. $\endgroup$ – voldemort Oct 17 '14 at 0:49

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