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I have problem with the divergence of a Sum, I don´t know how criteria use when i has a trigonometrical function because, it´s not monotone, and I can´t prove that this series diverge with the usual methods $$ \sum {\frac{{\cos \left( {\log \left( {\log n} \right)} \right)}} {{\log n}}} $$ it´s different

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This one is not very friendly, is it. – Patrick Da Silva Aug 9 '11 at 15:08
up vote 9 down vote accepted

Because the denominator varies so slowly and the numerator even more slowly, you can show that the sequence of partial sums $S_k$ can't be Cauchy by adding up a suitable stretch of terms. For instance, with $k=\lceil\exp(\exp(2\pi m))\rceil$ and $l=\lfloor \exp(2\exp(2\pi m))\rfloor$, we have

$$S_l-S_k=\sum_{j=k+1}^l\frac{\cos(\log(\log j))}{\log j}\gt\sum_{j=k+1}^l\frac{\cos(2\pi m+\log2)}{\log l}\gt(l-k)\frac{\cos\log2}{\log l}>\cos\log 2\;,$$

where the last inequality holds for sufficiently large $m$. Since the difference of partial sums doesn't go to zero, the series diverges.

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Great answer. +1 – Patrick Da Silva Aug 9 '11 at 15:20
Good answer, but i did not understand the first inequality D: only that. – Daniel Aug 9 '11 at 15:26
@Daniel: The denominators have increased. In the numerator, the arguments of the cosine lie between $2\pi m$ and $2\pi m + \log2$, corresponding to $0$ to $\log2$, and the values of the cosine in that range are $\ge\cos\log2$. – joriki Aug 9 '11 at 15:36
Ah ok I got it, very good answer, , i did not saw "l" in the denominator , thanks again :D – Daniel Aug 9 '11 at 15:39

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