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So I'm trying to prove some series converges, and I'm trying to show it by using the Drichlet test.

So I need to prove that the series of partial sums $\Sigma_{k=1}^\infty \sin(ki)$ is bounded. I tried proving it by dividing and multiplying with $2\cos(\frac{i}{2})$ and then using the trigonometric identity $2\cos(\beta)\sin(\alpha)=sin(\alpha+\beta)-sin(\alpha-\beta)$, which creates a telescopic series - but I didn't manage to bound the result.

Any assistance would be great! Thanks in advance!

Reyo
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3 Answers3

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One way to obtain a bound is to derive a closed-from expression for the sum. Here, this is accomplished by summing geometric series. To that end,

$$\begin{align} \left|\sum_{n=1}^N \sin(nx)\right| &=\left| \text{Im} \left(\sum_{n=1}^N e^{inx} \right) \right|\\ &=\left|\text{Im} \left( \frac{e^{ix}-e^{i(N+1)x}}{1-e^{ix}} \right)\right| \\ &=\left|\frac{\sin \left(\frac{Nx}{2} \right) \sin\left(\frac{(N+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right|\\ &\le \left|\csc\left(\frac{x}{2}\right)\right| \end{align}$$

Note: A tighter bound can be obtained by noting $$\left|\sin \left(\frac{Nx}{2} \right) \sin\left(\frac{(N+1)x}{2}\right)\right|= \frac12 \left||\cos (\frac{x}{2})-\cos(N+\frac12)x \right|\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)$$

Mark Viola
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I'll use $x$ in place of $i$ in order to avoid confusion with the imaginary number $\sqrt{-1}$. Also, I'll assume $x$ is not an integral multiple of $2\pi$.

Since

\begin{align}\sum_{k = 1}^n \sin(kx) &= \csc(x/2) \sum_{k = 1}^n \sin(kx)\sin(x/2)\\ & = \frac{1}{2}\csc(x/2) \sum_{k = 1}^n [\cos((k - 1/2)x) - \cos((k + 1/2)x)]\\ & = \frac{1}{2}\csc(x/2) [\cos(x/2) - \cos((n+1/2)x] \end{align}

and the cosine is bounded by $1$,

$$\left|\sum_{k = 1}^n \sin(kx)\right| \le |\csc(x/2)|.$$

Since the upper bound is independent of $n$, the sequence of partial sums of $\sum_{k = 1}^\infty \sin(kx)$ is bounded.

kobe
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You are almost there. Since:

$$\begin{eqnarray*} 2\sin\frac{i}{2}\sum_{k=1}^{K}\sin(ki) &=& \sum_{k=1}^{K}\left(\cos\left(\left(k-\frac{1}{2}\right)i\right)-\cos\left(\left(k+\frac{1}{2}\right)i\right)\right)\\&=&\cos\frac{i}{2}-\cos\frac{(2K+1)i}{2}&\end{eqnarray*}$$

we have that for any $K$: $$\left|\sum_{k=1}^{K}\sin(ki)\right|\leq\frac{\left|\cos\frac{i}{2}\right|+1}{2\left|\sin\frac{i}{2}\right|}$$ This bound is optimal since for any $\varepsilon > 0$ we can find a $K\in\mathbb{N}$ such that the difference between the LHS and the RHS of the last line is less than $\varepsilon$.

Jack D'Aurizio
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  • It is correct, also checked with WA. It is just a stronger inequality than yours, since you have bounded $\left|\cos\frac{i}{2}\right|$ with one, but there is no need of doing that. – Jack D'Aurizio Mar 25 '15 at 22:33
  • Well, I didn't enforce that particular bound explicitly, but implicitly only. Rather, I enforced $|sin(Nx/2)sin((N+1)x/2)|\le1$. Of course, the term under the absolute value can be written $\frac{1}{2}|\cos(x/2)-\cos((N+1/2)x)|$. So, I bounded this entire expression by $1$ by tacitly bounding both terms under the absolute value by $1$. – Mark Viola Mar 25 '15 at 22:56
  • I tried a numerical check. Suppose $K=1$ and $i=5$. Then $|\sin(5)| \approx 0.95892$ while $\frac12 \cot(5/4)| \approx 0.1661$. I believe that the problem is with the last step in which you applied an identity by removing absolute values and then reapplying them thereafter. – Mark Viola Mar 25 '15 at 23:16
  • @Dr.MV: ok, you're right. Fixed by just removing the very last piece. – Jack D'Aurizio Mar 26 '15 at 00:27
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    You're a very creative mathematician and I always appreciate reading your solutions. Just glad I could help for once! – Mark Viola Mar 26 '15 at 02:38