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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.

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What is $ec$?.. –  leo Feb 23 '12 at 20:12
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@leo I guess Hardy meant $\operatorname{cosec}$. –  Davide Giraudo Feb 23 '12 at 20:20

2 Answers 2

up vote 2 down vote accepted

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.

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i do n't follow the first line in your solution how is $\left|\sum_{j=1}^{N+n}f_j(\theta)\sin(j\theta)-\sum_{j=1}^{N+n}f_j(\theta)\sin(‌​j\theta)\right|=\left|\sum_{j=N+1}^{N+n}f_j(\theta)\sin(j\theta)\right|$ ? –  Hardy Feb 23 '12 at 22:41
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@Hardy There was a typo, hope it's fixed now. –  Davide Giraudo Feb 27 '12 at 19:57
    
Thanks very much for that. Just one last thing how did u know that Abel's transform is needed here. This is so i can recognize such problems myself. –  Hardy Feb 27 '12 at 20:03
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We have a sum of the form $\sum_k a_k b_k$, we know that the sequence $\{s_n=\sum_{k=1}^nb_k\}$ is bounded and we know that $a_k$ is monotone, so after Abel's transform we will have a sum of the form $\sum_k (a_k-a_{k+1}s_k$. The last sum involves terms on which we have hypothesis. –  Davide Giraudo Feb 27 '12 at 20:09

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 $$\sum_{n=1}^{+\infty} |f_{n+1}-f_n| \cdot |S_n| \leq \sup_n(S_n) \sum_{n=1}^{+\infty} (f_n-f_{n+1}) = \sup_n(S_n) f_1.$$

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Firstly thank you very much for your reply i do think it is easier for me to follow. I suspect though there is a mistake somewhere as it seems to me that you proved this for all values of $\theta$ where as the question states it should only converge when $\theta$ is not even multiple of $\pi$. Am i overlooking or misunderstanding anything here. –  Hardy Feb 27 '12 at 19:34
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$\theta \mod 2\pi \neq 0$ involve the series with $\cos$. I let you deal with this case. –  user10676 Feb 27 '12 at 19:49

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