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$$\sum_{n=1}^\infty\frac{\sin n}{\ln n+\cos n}$$ My guess is "yes", but I can't prove it.

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Why is your guess "yes"? – Pedro Tamaroff Jun 20 '14 at 21:07
I rather like this question, but I've no idea how to approach it. A naive Dirichlet's test is hopeless because of the $\cos n$ in the denominator flipping the signs all over. Partial summation doesn't seem to get me anywhere meaningful. I don't even know whether I think it converges or diverges – mixedmath Jun 20 '14 at 21:30
I think it's reasonable to guess "yes" because the series $(-1)^n/\log{n}$ converges. That said, I too cannot think off the top of my head how to approach this. – Ron Gordon Jun 20 '14 at 21:41
Pedro Tamaroff, my heuristic argument is the following. It's not hard to show that the series $\sum\frac{\sin n}{\ln n+\sin n}$ diverges, since $\frac{\sin n}{\ln n+\sin n}=\frac{\sin n}{\ln n}-\frac{\sin^2n}{\ln^2n}(1+o(1))$. If you try to get such an expansion for the series in question then there never appear "diverging" term. In other words, in the second series $\sin n$ terms resonate while in the first $\sin n$ and $\cos n$ terms in the numerator and denominator have "phase shift". Of course that doesn't prove anything, so I'm asking for an idea of rigorous proof. – CuriousGuest Jun 20 '14 at 22:14
Break the series up by the sign of sin(n) so it looks like (-1)^n a_n and use the alternating series test. My initial guess is that the sum of sin(n) where n lies in any connected interval of [k(\pi),(k+1) \pi] should be bound by something like 4 and so I expect the series to be eventually decreasing. This is the same style of thought Ron brought forward. – Daniel Parry Jun 21 '14 at 1:19
up vote 18 down vote accepted

This method was inspired by the OP's heuristic argument in a comment to the question. We approximate the summand by Taylor polynomials, but with more and more terms as $n$ grows.

We need the fact that for every integer $k\ge1$, $$ \bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| = O(k^7) $$ uniformly for integers $B<C$. To see this, verify that $\sin n \cdot \cos^k n = q_k(e^{in})$, where $q_k(t)=\sum_{j=-k-1}^{k+1} c_{k,j}t^j$ is a Laurent polynomial with no constant term that satisfies $\sum_{j=-k-1}^{k+1} |c_{k,j}| = 2^{-k}\binom{k}{\lfloor k/2\rfloor} \le 1$. Then \begin{align*} \sum_{B\le n< C} \sin n \cdot \cos^k n &= \sum_{B\le n< C} q_k(e^{in}) \\ &= \sum_{j=-k-1}^{k+1} \sum_{B\le n< C} c_{k,j}e^{ijn} = \sum_{j=-k-1}^{k+1} c_{k,j} \frac{e^{ijC}-e^{ijB}}{e^{ij}-1}, \end{align*} so that $$ \bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| \le \sum_{j=-k-1}^{k+1} |c_{k,j}| \frac{2}{|e^{ij}-1|}. $$ The fact that $\pi$ has an irrationality measure of less than $8$ (Salikhov, 2012) means that $|e^{ij}-1|\gg j^{-7}$. Therefore $$ \bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| \ll k^7 \sum_{j=-k-1}^{k+1} |c_{k,j}| \ll k^7. $$

Also note that for any positive integer $A$ and any real number $x\in[-\frac12,\frac12]$, $$ \frac1{1+x} = \sum_{k=0}^{A-1} (-x)^k + r(x) $$ where $|r(x)| \le |2x|^A$ (a consequence of Taylor's theorem).

Now we establish the convergence of the series in question by showing that it is Cauchy, i.e., that $$ \sum_{n=M}^N \frac{\sin n}{\log n+\cos n} \to 0 $$ as $M\to\infty$ (and $N>M>e^{2e^2}$, say). For each $n$, we choose $x=\frac{\cos n}{\log n}$ and $A=\lceil{\log n}\rceil$, giving $$ \sum_{n=M}^N \frac{\sin n}{\log n} \bigg( \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k + s(n) \bigg) $$ where $|s(n)| \le |2\frac{\cos n}{\log n}|^A \le (\frac2{\log n})^{\log n} \le \frac1{n^2}$ for $n>e^{2e^2}$. Therefore the contribution from $\sum_{n=M}^N \frac{\sin n}{\log n}s(n)$ is $O(\frac1M)$. As for the other terms, we can write \begin{multline*} \sum_{n=M}^N \frac{\sin n}{\log n} \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k = \sum_{k<\log M} (-1)^k \sum_{n=M}^N \frac{\sin n \cdot \cos^k n}{\log^{k+1} n} \\ + \sum_{\log M<k<\log N} (-1)^k \sum_{e^k < n \le N} \frac{\sin n \cdot \cos^k n}{\log^{k+1} n}. \end{multline*}

Let $S_k(t) = \sum_{n\le t} \sin n \cdot \cos^k n$, which is $O(k^7)$ as noted above. Then \begin{align*} \sum_{B<n\le C} \frac{\sin n \cdot \cos^k n}{\log^{k+1} n} = \int_B^C \frac{dS_k(t)}{\log^{k+1} t} &= \frac{S_k(t)}{\log^{k+1} t} \bigg|_B^C + (k+1) \int_B^C \frac{S_k(t)}{t\log^{k+2} t}\,dt \\ &\ll k^7 \bigg(\frac1{\log^{k+1} B} + (k+1) \int_B^C \frac1{t\log^{k+2} t}\,dt \bigg) \\ &\ll k^7 \frac1{\log^{k+1} B}. \end{align*} In particular, \begin{align*} \sum_{n=M}^N \frac{\sin n}{\log n} \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k &\ll \sum_{k<\log M} \frac{k^7}{\log^{k+1} M} + \sum_{\log M<k<\log N} \frac{k^7}{\log^{k+1} e^k} \\ &\ll \sum_{k=1}^\infty \frac{k^7}{\log^{k+1} M} + \sum_{k=2}^\infty \frac1{k^{\log M-6}}, \end{align*} and both these series tend to $0$ as $M\to\infty$ by the dominated convergence theorem.

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Great proof, thank you! – CuriousGuest Jun 21 '14 at 8:43

Why not just write $$\frac{\sin n}{\log n + \cos n} = - \frac{\sin n}{\log n } \left(1- \frac{1}{1 + \frac{\cos n}{\log n }} \right) + \frac{\sin n}{\log n } $$ The first term gives an absolutely convergent series, since $$ \left| \frac{\sin n}{\log n } \left(1- \frac{1}{1 + \frac{\cos n}{\log n }} \right) \right| \sim \left| \frac{\sin 2n}{2 \log^2 n} \right| \leq \frac{1}{2 \log^2 n} $$ And for the second term, the Dirichlet criterion applies because $\frac{1}{\log n}$ is decreasing to $0$. Thus the series converges.

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$$\sum_{n=2}^\infty \frac{1}{\log^2 n} = \infty,$$ the estimate does not converge, and the series with term $\frac{\sin (2n)}{2\log^2n}$ most likely does not converge absolutely. – Daniel Fischer Jun 27 '14 at 21:13

This is not an answer, but some computational "evidence". Below is a plot of the values of the sum for up to $20.000$ summations. As can be seen in the picture, the sum oscillates for a very long time, but one might guess that the oscillations die away because of the $\ln n$-term, after a very long time. What can be concluded however, is that if the answer is "no", then it is not because the sum tends to infinity, but because of the oscillations.

enter image description here

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The terms tend to zero so I'm not sure an oscillating limit is possible, I mean $\log(20,000)$ is still less than $10$. – John Fernley Jun 20 '14 at 23:04
I have to (mostly) agree with John here - for series with terms of this magnitude, you'd need millions if not hundreds of millions of terms just to get a couple digits' worth of accuracy; this plot is on too small a scale to draw even tentative conclusions from. – Steven Stadnicki Jun 21 '14 at 0:47
@StevenStadnicki I agree with both of you. However, as stated, my answer is not an answer, and I believe the picture gives good insight into why this is a difficult sum to evaluate. – Fredrik Meyer Jun 21 '14 at 7:01

The sum $ | \sum_{n=3}^{\infty} \frac{\sin(n)}{\ln(n)-1}|$ converges by the Dirichlet-Criterion since $ | \sum_{n=3}^{\infty} sin(n)|$ is bounded and $a_n = \frac{1}{\ln(n)-1}$ is a decreasing positive sequence converging to zero. Then it follows that $$ \sum_{n=3}^{\infty} \frac{\sin(n)}{\ln(n)+\cos(n)} < \sum_{n=3}^{\infty} \frac{\sin(n)}{\ln(n)-1} < \infty$$

Surely that would be too easy though, but where is my mistake? :)

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Since the terms have different signs, $\lvert a_n\rvert \leqslant \lvert b_n\rvert$ does not imply that $\sum a_n$ converges, given that $\sum b_n$ converges. Take for example $b_n = \frac{(-1)^{n+1}}{n}$ and $a_n = b_n$ if $n$ is odd, $a_n = \frac{1}{n^2}$ if $n$ is even. – Daniel Fischer Jun 30 '14 at 14:04

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