Show $\sum _{n=1}^{\infty }f_{n}\cos \left( n\theta \right) $ converges  I am having trouble proving the third part of the problem.
"Show if $0 < \theta < 2\pi $, $\left| \sum _{n=1}^{p}\sin \left( n\theta \right) \right| < \cos ec\dfrac {\theta } {2}$; and deduce that, if $f_{n}\rightarrow 0$ steadily(synonym for monotonically), $\sum _{n=1}^{\infty }f_{n}\sin \left( n\theta \right) $ converges for all real values of $\theta$, and that $\sum _{n=1}^{\infty }f_{n}\cos \left( n\theta \right) $ converges if $\theta$ is not even multiple of $\pi$.
Here is where i got upto, to see that the inequality holds i used Lagrange's trigonometric identity in the LHS and the result follows.
Once that is assumed to hold, $\sum _{n=1}^{\infty }f_{n}\sin \left( n\theta \right) $ just follows from  Dirichlet's test of convergence. 
I am unsure how to show the last one i assume the problem is some how related to Abel's inequality or Dirichlet's test of convergence, any help would be much appreciated.
 A: We can use Abel's transform: for $N\in\mathbb N$ and $n\geq 0$ we have, denoting $s_j(\theta):=\sum_{k=0}^j\sin(n\theta)$
\begin{align*}
\left|\sum_{j=1}^{N+n}f_j(\theta)\sin(j\theta)-\sum_{j=1}^Nf_j(\theta)\sin(j\theta)\right|&=\left|\sum_{j=N+1}^{N+n}f_j(\theta)\sin(j\theta)\right|\\
&=\left|\sum_{j=N+1}^{N+n}f_j(\theta)(s_j(\theta)-s_{j-1}(\theta))\right|\\
&=\left|\sum_{k=N+1}^{N+n}f_k(\theta)s_k(\theta)-\sum_{k=N}^{N+n-1}f_{k+1}(\theta)s_k(\theta)\right|\\
&\leq \left|f_{N+1}(\theta)s_N(\theta)\right|+\left|f_{N+n}(\theta)s_{N+n}(\theta)\right|\\
&+\sum_{j=N+1}^{N+n}\left|(f_k(\theta)-f_{k+1}(\theta))s_k(\theta\right|\\
&\leq \frac 1{|\sin\frac {\theta}2|}(|f_{N+n}(\theta)|+|f_{N+1}(\theta)|\\
&+\sum_{j=N+1}^{N+n}|f_k(\theta)-f_{k+1}(\theta)|),
\end{align*}
and now the monotone convergence to $0$ of $f_n(\theta)$ gives the result.
A: As pointed by Davide Giraudo, the key is Abel's transform. I'll try to write an answer clearer for you.
Fix $\theta \in \mathbb{R}$. Let $S_n = \sum_{k=1}^n \sin(n\theta)$ for all $n \in \mathbb{N}$. Using Abel transform you have, for $p \in \mathbb{N}_{\geq 1}$ :
$$\sum_{n=1}^p f_n \sin(n\theta) = \sum_{n=0}^p f_n (S_n-S_{n-1}) = (f_p S_p - f_1 S_0) - \sum_{n=1}^{p-1} (f_{n+1}-f_n) S_n.$$
Since you proved that $(S_p)_p$ is bounded, you have
$$\lim_{p\to +\infty} f_p S_p=0.$$
Now we show that the $\sum_{n=1}^{+\infty} (f_{n+1}-f_n) S_n$ is absolutely convergent. This is because
$$\sum_{n=1}^{+\infty} |f_{n+1}-f_n| \cdot |S_n| \leq \sup_n(S_n) \sum_{n=1}^{+\infty} (f_n-f_{n+1}) = \sup_n(S_n) f_1.$$
