Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

share|cite|improve this answer
i do n't follow the first line in your solution how is $\left|\sum_{j=1}^{N+n}f_j(\theta)\sin(j\theta)-\sum_{j=1}^{N+n}f_j(\theta)\sin(‌​j\theta)\right|=\left|\sum_{j=N+1}^{N+n}f_j(\theta)\sin(j\theta)\right|$ ? – Comic Book Guy Feb 23 '12 at 22:41
@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. – Comic Book Guy Feb 27 '12 at 20:03
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 $$\sum_{n=1}^{+\infty} |f_{n+1}-f_n| \cdot |S_n| \leq \sup_n(S_n) \sum_{n=1}^{+\infty} (f_n-f_{n+1}) = \sup_n(S_n) f_1.$$

share|cite|improve this answer
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. – Comic Book Guy Feb 27 '12 at 19:34
$\theta \mod 2\pi \neq 0$ involve the series with $\cos$. I let you deal with this case. – user10676 Feb 27 '12 at 19:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.