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I am studying $\sum_{n\neq0} \frac{e^{inx}}{n}$ better known as the saw-tooth function.

In order to apply Dirichlet's test for convergence, which states:

If the partial sums of the series $\sum b_n$ are bounded, and {$a_n$} decreases monotonically to 0, then $\sum a_nb_n$ converges.

I need that $\sum e^{inx} $ is bounded, but as geometric series with r=1, things are more complicated. Also if we take $x=2\pi$, then it is unbounded.

Additionally, it is my understanding that these indices begin at 1 and run through the natural numbers, and I am skeptical of if this theorem can be applied from a sum from -N to N and the like with complex series.

Thank you, and hints are greatly appreciated.

Jungleshrimp
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1 Answers1

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Note that we have for $x\ne 0, \pm 2\pi, \pm 4\pi, \pm 6\pi \dots$,

$$\sum_{n=1}^N e^{inx}=\frac{e^{i(N+1)x}-e^{ix}}{e^{ix}-1}=e^{i(N+1)x/2}\frac{\sin(Nx/2)}{\sin(x/2)}\tag1$$

Hence for $x\ne 2k\pi$ $k\in \mathbb{Z}$, the term on the right-hand side of $(1)$ is bounded in magnitude by $|\csc(x/2)|$.

Also, note that for $x\ne 2k\pi$ $k\in \mathbb{Z}$

$$\sum_{n=-N}^{-1} e^{inx}=e^{-i(N+1)x/2}\frac{\sin(Nx/2)}{\sin(x/2)}$$

which has the same bound as $(1)$.

Mark Viola
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