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The title is the question: I'm trying to understand the behavior of the series

$$\star \qquad \sum_{n=1}^\infty \frac{(-1)^n \cos(n^2 x)}{n},\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad $$

where $0\leq x \leq 1$. All I know for sure about it is that it converges conditionally at $x=0$ to a known value (I think it's $-\ln 2$). I don't even know how to check its convergence for other values of $x$ (it may be possible to compute it for certain rational multiples of $\pi$). I plotted the partial sum going up to $n=1000$ using Maple and the partial sum only goes from about $-2.8$ to $0.4$. It also looks continuous but very jagged. This suggests that the series may converge at least pointwise to a continuous, nowhere-differntiable function. The maximum value of the absolute value of the derivative of the 1000th partial sum is about $21249$, according to Maple17's Mamimize command, but I am suspicious because Maple says the maximum occurs at $x=0.2000002655$, suspiciously close to $0.2$. But if the derivative of the 1000th partial sum really is that big, it suggests that if the series converges to a function, that function might be nowhere-continuous.

I did some Googling, looking for information on continuous, nowhere-differentiable functions, hoping to find my series. There seem to be a huge number of them that are well-known. The best-known such functions seem to be "Weierstrass functions", but for such functions the denominator is something like $2^n$, which grows much faster than $n$. At http://epubl.luth.se/1402-1617/2003/320/LTU-EX-03320-SE.pdf , there is posted a thesis about continuous, nowhere-differentiable functions. On p. 23 of the .pdf (not the author's page number, but the page number of the .pdf document), the author discusses "Riemann's function", defined by $\sum_{k=1}^\infty \frac{1}{k^2}\sin(k^2 x)$. This is very similar to my series, but again the denominator decays more rapidly than the one I'm considering.

So, to make a long story short, this is what I'd like to know:

For what values of $x$ does $\star$ converge?

For what values of $x$ does $\star$ converge absolutely?

If $\star$ converges for all $x$ (EDIT: NO: see accepted answer) , does it converge merely pointwise or uniformly? (see answer)

If $\star$ converges at least pointwise (EDIT: NO: see accepted answer), where is the resulting function differentiable?

EDIT: I accepted an excellent answer that answers most of my questions, and I will not unaccept it. However, I am still a little curious about the complete answers to my first and second questions above. I will upvote informative, original comments and answers.

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$$(-1)^n\cos (n^2x)=cos(n^2x-n\pi)$$ –  ProMatheus Nov 24 '13 at 4:01
1  
Oh, and the convergence at $x=0$ is not absolute, since the absolute value of the terms is just the harmonic series. –  Henning Makholm Nov 24 '13 at 4:15
    
Why are you saying the series converges absolutely for $x=0$? It converges conditionally. –  Mhenni Benghorbal Nov 24 '13 at 5:09
    
@SurajM.S : Your observation is useful if one wants to repeat Henning's argument for other rational multiples of $\pi$. –  Stefan Smith Nov 29 '13 at 0:43

1 Answer 1

up vote 3 down vote accepted

The series diverges for $x=\frac{2\pi}8\approx 0.78$. Since the squares modulo 8 are 0,1,4,1,0,1,4,1, the series for this $x$ becomes $$ - \frac{\sqrt{1/2}}{1} - \frac1{2} - \frac{\sqrt{1/2}}{3} + \frac1{4} - \frac{\sqrt{1/2}}{5} - \frac1{6} - \frac{\sqrt{1/2}}{7} + \cdots$$

The terms with even denominator form an alternating series -- their $(-1)^n$ factor is always $1$ but the cosine alternates between $-1$ and $1$. So their sum converges. However, the terms with odd denominator all have the same sign because the $(-1)^n$ factor is always $-1$ and the cosine is always $\cos\frac{\pi}{4} = \sqrt{1/2}$. So they diverge logarithmically towards $-\infty$.

I would expect (without having checked) that a similar divergence will happen at many other rational multiples of $\pi$, such that your series diverges for a dense set of $x$s.

Note that logarithmic divergence is slow. Even for $1000$ terms, you shouldn't expect it to have reached more than $\log 1000\approx 7$ times the first term, so the $2.8$ maximum you've found numerically does not really point towards convergence (note that in the above case only half of the terms participate in the divergence).

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Again, that is excellent, and it shows that computers can't replace analysis. I am also quite sure that the series will diverge for a set of rational multiples of $\pi$ that is dense in $[0,1]$. –  Stefan Smith Nov 24 '13 at 4:09

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