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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.

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@All, I am "hearing crickets" here boys, or should i say "reading crickets" :-). but in all sincerity, please do let me know if there is anything i could do get some one else interested enough to consider my problem and possibly offer a hint. –  Hardy Mar 25 '12 at 7:42
    
What do you mean by $u_n\to0$ "steadily"? –  Julián Aguirre Mar 25 '12 at 10:42
    
@JuliánAguirre it means $u_n\rightarrow 0$ monotonically. –  Hardy Mar 25 '12 at 17:47
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2 Answers

up vote 3 down vote accepted
+50

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.

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does n't the boundedness of the partial sums under the conditions u showed along with G.H.hardy's result above show uniform convergence ? also why does the ratio test need convergence of infinite sum of $U_n$ ? –  Hardy Mar 25 '12 at 20:44
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You will get uniform convergence on any compact subset of $(0,2\,\pi)$. I do not understand your question about the ratio test. If $\sum u_n<\infty$, the series would converge uniformly absolutely; you can check if this is so applying the ratio test. But you have no information at all tat allows you to prove that $\limsup u_{n+1}/u_{n}<1$, much less that $\lim u_{n+1}/u_{n}<1$. All you know for sure is that $\limsup u_{n+1}/u_{n}\le1$, but this is not enough to obtain convergence of $\sum u_n$. –  Julián Aguirre Mar 25 '12 at 22:07
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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.

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I thought about using Dirichlet's test but i believe it can not be used in this case because of the existence of the $a$ inside the parameter of $cos$. Also i could be wrong about the assumption that $\lim_n\to\infty|u_{n+1}/u_n|<1$ but there is a theorem about convergence $\lim_n\to\infty|u_{n+1}/u_n|=1$ too so perhaps since $u_n$ is monotonic we can exhaust the two options that way. –  Hardy Mar 25 '12 at 18:52
    
it appears buddy that u were right about the Dirichlet's test usage after all. –  Hardy Mar 25 '12 at 19:34
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