Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 1 / 9D (Analysis I)

Use $\sin 0=0$, the derivative identity ${\sin }^{\prime }=\cos$, continuity of cosine at zero, and $\cos 0=1$. For $u\ne 0$, the mean value theorem gives $\sin u/u=\cos c_u$ for a point between zero and $u$. Hence $\sin u/u\to 1$ as $u\to 0$. For $x\ne 0$, apply this with $u=x/3^k$ to obtain the limit $\boxed{3^k\sin (x/3^k)\to x}$. For $x=0$ every term is zero.

The additional bound $|\cos t|\le 1$ and the same mean value theorem give $|\sin u|\le |u|$. ThereforeThe right side is summable. By the comparison test for series the stated series has absolute convergence for every real $x$.

Let $r=e^{i\theta }\ne 1$. The geometric series from any starting index is bounded:Using summation by parts,Since the coefficients are nonincreasing and positive, the modulus is at most $K(a_N+a_m-a_N)=K{a}_m\to 0$. The Cauchy criterion proves convergence for every $\theta \notin 2\pi \mathbb{Z}$. This is a direct proof of the Dirichlet test in this setting.

For $a_n=1/n$, the imaginary parts of the convergent partial sums give convergence of the sine series at those angles. If $\theta \in 2\pi \mathbb{Z}$, every sine term is zero. Hence the sine series converges for all real angles.