proving or refuting the convergence of a digital sequence

Question

Let the following digital sequence; $\Sigma_{n=2}^\infty \dfrac{sin(nx)}{log(n)}$

Dirichlet's criteria says that if $b_n$ decreases and $lim$ $b_n =0$ and if the partial sums of a sequence $a_n \in \mathbb{R}$ then $\Sigma_{n=1}^\infty a_n b_n$ converges.

• It's obvious that $b_n$ in this case is $1/log(n)$ and it does indeed decrease and converges in $0$.

• Now what I have trouble on is showing that all the partial sums of $a_n$ (in this case $sin(nx)$) converge. I know that showing that all the partial sums of a sequence converge is equivalent to showing that $\exists M>0, \forall n \vert{S_n}\vert \leq M$ With $S_n := \Sigma_{n=1}^\infty a_n$

I found a proof (that all the partial sums of $a_n$ converge) but I can't seem to understand the first step. It goes like this:

$\vert \vert \Sigma_{k=1}^\infty sin(kx) \vert \vert$ = $\vert \vert \dfrac{sin(\dfrac{1}{2} nx) sin ((n+1)\dfrac{x}{2})}{sin\dfrac{x}{2}} \vert \vert$

I really don't know where that result come from

Answer 1

In order to apply the Dirichlet's Test you still have show that the sequence $\{a_n\}_{n\geq 2}$ is bounded.

Now by the addition formula $\cos(x+y)=\cos x \cos y -\sin x \sin y$, for $x\not=2 m\pi$ with $m\in\mathbb{Z}$ (otherwise $a_n=0$), it is easy to obtain $$\sin(kx)=\frac{\cos\bigl((k-\frac{1}{2})x\bigr)-\cos\bigl((k+\frac{1}{2})x\bigr)}{2\sin\frac{x}{2}}.$$ Hence $$\sum_{k=1}^{n} \sin(kx)=\frac{1}{2\sin\frac{x}{2}}\sum_{k=1}^n \left[\cos\bigl((k-\frac{1}{2})x\bigr)-\cos\bigl((k+\frac{1}{2})x\bigr)\right]=\frac{\cos\bigl((1-\frac{1}{2})x\bigr)-\cos\bigl((n+\frac{1}{2})x\bigr)}{2\sin\frac{x}{2}}$$ and $$|a_n|=\left |\sum_{k=1}^{n} \sin(kx)=\right|\leq \frac{1}{\left|\sin\frac{x}{2}\right|}.$$ So the $\{a_n\}_{n\geq 2}$ is bounded and by the Dirichlet's Test, your series is convergent.

Answer 2

let the statement be true for $k=n-1$ we show it is also true for $k=n$

$\Sigma_{k=1}^n \sin(kx) \sin\frac{x}{2}$

=$\Sigma_{k=1}^{n-1} \sin(kx) \sin\frac{x}{2}+ \sin(nx) \sin\frac{x}{2}$

= $\sin(\frac{1}{2} nx)\{ \sin ((n-1)\frac{x}{2})+2 \cos\frac{nx}{2} \sin\frac{x}{2}\}$

=$\sin(\frac{1}{2} nx)\{ \sin ((n-1)\frac{x}{2})+\sin ((n+1)\frac{x}{2})-\sin ((n-1)\frac{x}{2})\}$

=$\sin(\frac{1}{2} nx) \sin ((n+1)\frac{x}{2})$

Answer 3

Another way is to observe that $$\sum_{k=1}^{n}\sin\left(kx\right)=\textrm{Im}\left(\sum_{k=1}^{n}e^{ikx}\right) $$ and the last sum is easy to evaluate $$\sum_{k=1}^{n}e^{ikx}=\frac{e^{ix}\left(1-e^{inx}\right)}{1-e^{ix}}=\frac{\left(\cos\left(x\right)+i\sin\left(x\right)\right)\left(1-\cos\left(nx\right)-i\sin\left(nx\right)\right)}{1-\cos\left(x\right)-i\sin\left(x\right)} $$ and now if we multiply the numerator and the denominator by $1-\cos\left(x\right)+i\sin\left(x\right) $ and we do the multiplications we can find the imaginary part.