# Does the series $\sum_{n=1}^\infty\frac{\sin n}{\ln n+\cos n}$ converge?

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

• 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 – davidlowryduda 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

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.

• 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.

• $$\sum_{n=2}^\infty \frac{1}{\log^2 n} = \infty,$$ 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.

• 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? :)

• 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