Behaviour of $\sum u_{n}\cos(n\theta +a) $ when $\theta\over\pi$  is rational and when irrational I am trying to show that if $\theta$ is not equal to $0$ or a multiple of $2\pi$, and if $u_0,u_1,u_2,...$ be a sequence such that $u_n\rightarrow 0$ steadily, then the series $\sum u_{n}\cos(n\theta +a) $ is convergent and also if the limit of $u_{n}$ is not zero, but $u_{n}$ is still monotonic, the sum of the series is oscillatory if $\theta\over\pi$ is rational, but that, if $\theta\over\pi$ is irrational, the sum may have any value between certain bounds whose difference is $\lim _{n\rightarrow \infty }u_{n}cosec\dfrac {\theta } {2}$. 

Solution  Attempt
Based on the upper bound of the partial sums are bounded as shown by Julián Aguirre
below and based on G.H.Hardy's test for uniform convergence which states that if, in a given domain, $\left| \sum _{n=1}^{p}a_{n}\left( z\right) \right| \leq k$ where $a_{n}\left(z\right)$ is real and $k$ is finite and independent of p and z, and if $f_{n}\left( z\right) \geq f_{n+1}\left( z\right) $ and $f_{n}\left( z\right)\rightarrow 0$ uniformly as $n\rightarrow \infty$, then $\sum _{n=1}^{\infty }a_{n}\left( z\right) f_{n}\left( z\right) $ converges uniformly.
Since $u_n\rightarrow 0$ and the sequence $u_0,u_1,u_2,...$ is monotonically decreasing. 
We also observed that the  partial sum is bounded so the series converges uniformly. 
Hence the series must converge.
I am unsure how to attack the second part of the problem, to be more specific how the cos function varies when the parameter part is irrational. Any help would be much appreciated.
 A: The hypothesis do not say anything about the convergence of the series $\sum u_n$. There is nothing you can say about the absolute convergence of the series, and in particular, the ratio test is useless.
The series converges by Dirichlet's test. It is enough to show that the partial sums $\sum_{k=1}^n\cos(n\,\theta+a)$ are bounded for $\theta\ne2\,k\,\pi$:
$$
\sum_{k=1}^n\cos(n\,\theta+a)=\Re\Bigl(\sum_{k=1}^ne^{(n\,\theta+a)i}\Bigr)=\Re\Bigl(e^{ai}\sum_{k=1}^ne^{n\theta i}\Bigr)=\Re\Bigl(e^{ai}\frac{e^{(n+1)\theta i}}{e^{\theta i}-1}\Bigr),
$$
$$
\Bigl|\sum_{k=1}^n\cos(n\,\theta+a)\Bigr|\le\frac{1}{|e^{\theta i}-1|}.
$$
For the second question, if $\lim_{n\to\infty}u_n=u$, then $u_n-u$ converges monotonically to $0$. Writing
$$
\sum_{k=1}^nu_n\cos(n\,\theta+a)=u\sum_{k=1}^n\cos(n\,\theta+a)+\sum_{k=1}^n(u_n-u)\cos(n\,\theta+a).
$$
By the previous argument, $\sum_{k=1}^n(u_n-u)\cos(n\,\theta+a)$ converges; let $S$ be its sum. We see that it is enough to study the behaviour of
$$
\sum_{k=1}^n\cos(n\,\theta+a)=\Re\Bigl(e^{ai}\frac{e^{(n+1)\theta i}}{e^{\theta i}-1}\Bigr)=\frac12\csc\frac\theta2\sin\Bigl(\bigl(n+\frac12\bigr)\theta+a\Bigr).
$$
If $\theta/\pi$ is irrational, then the values of
$$
\sin\Bigl(\bigl(n+\frac12\bigr)\theta+a\Bigr)=\sin\Bigl(\frac{(2\,n+1)\pi}{2}\frac{\theta}{\pi}+a\Bigr)
$$
are uniformly distributed in $[-1,1]$, and the values of $\sum_{k=1}^n\cos(n\,\theta+a)$ are uniformly distributed in
$$
\Bigl[S-\frac{u}2\csc\frac\theta2,S+\frac{u}2\csc\frac\theta2\Bigr].
$$
On the other hand, if $\theta/\pi=p/q$ is rational, then
$$
\sin\Bigl(\bigl(n+\frac12\bigr)\theta+a\Bigr)=\sin\Bigl(\frac{(2\,n+1)p\pi}{2\,q}+a\Bigr)
$$
is a periodic sequence.
A: You cannot use the Ratio Test because you don't know $\lim_{n\to\infty}|u_{n+1}/u_n|<1$. The first part of your problem follows from Dirichlet's Test.
