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I'm trying to find convergence or divergence of $$\sum_{n=1}^\infty\frac{\cos(\log n)}{n}\text{.}$$I've tried to use the squeeze theorem, however, my professor said that $\log(n)$ grows too slowly for the numerator to be assumed to be zero. I don't see a solution with nth term, limit comparison or direct comparison. Any suggestions would be great.

Jack
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Justin Clark
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1 Answers1

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The series does not converge because it oscillates too much. Let $$S_m = \sum_{n=1}^m \frac{\cos \ln n}{n}$$ be the sequence of partial sums. Then for arbitrary $N > 0$, we can find $m_1, m_2$ with $|S_{m_1} - S_{m_2}| > 1$, so the partial sums continue oscillating by a constant amount no matter how far out you go.

(In real analysis terms, $(S_m)$ does not converge because it is not Cauchy.)

To prove this, given any $N>0$, choose $k$ such that $e^{(k-\frac13)\pi} > N$ and consider the sum $$\sum_{n = \exp((k-\frac13)\pi}^{\exp((k+\frac13)\pi)} \frac{\cos \ln n}{n}.$$ On this interval, we have $(k - \frac13)\pi \le \ln n \le (k + \frac13)\pi$ and therefore either $\cos \ln n \ge \frac12$ or $\cos \ln n < -\frac12$ according to whether $k$ is even or odd. The two cases are symmetric, so let's assume $k$ is even.

In that case, each term of the sum is at least $\frac{1}{2n}$, and we have $$\sum_{n=a}^b \frac{1}{2n} \approx \frac12 \left(\ln b - \ln a\right)$$ with an error that goes to $0$ as we choose large enough $a, b$. (We know this from an asymptotic estimate of the harmonic numbers (Wikipedia) or from a comparison to the integral of $\frac{1}{2x}$ between $a$ and $b$.)

In this case, $a = \exp((k-\frac13)\pi)$ and $b = \exp((k+\frac13)\pi)$, so $\frac12(\ln b - \ln a) = \frac\pi3$. If $k$ is chosen so that our estimates are precise enough (and we can choose $k$ to be as large as we want) then $$\sum_{n = \exp((k-\frac13)\pi}^{\exp((k+\frac13)\pi)} \frac{\cos \ln n}{n} > \frac\pi3 - 0.001 > 1.$$

Misha Lavrov
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  • This is a nice answer, but no calcII student is going to figure this out. Would the intergral test work? Setting $lnn=t$, you are then integrating $cost$ from 0 to infinity: Divergent. – imranfat Apr 15 '17 at 00:02
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    @imranfat Monotonic? Nope, integral test does not apply. – Simply Beautiful Art Apr 15 '17 at 00:03
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    The integral test doesn't apply because $\frac{\cos \ln n}{n}$ is neither nonnegative nor monotone decreasing. This answer is essentially applying the integral test on large blocks of the sequence that are nonnegative, however. – Misha Lavrov Apr 15 '17 at 00:03
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    @imranfat You can try to improve the readibility, but there is no easier way, than saying : $\cos \ln n > c$ for $\ln n \in (2k-\pi/3+c,k-2\pi/2-3) \implies n \in [e^{\pi(2k-1/3)},e^{\pi(2k+1/3)}]$ – reuns Apr 15 '17 at 00:04
  • @MishaLavrov Yes correct, I remember that now regarding the integral test. The readability of your answer is fine (also +1), but the OP mentioned this as a CalcII problem, and I cannot find any method that works within the framework of the course to tackle this problem – imranfat Apr 15 '17 at 00:06
  • @SimplyBeautifulArt. Yes, I remember now...But how to do this one with standard CalcII stuff? – imranfat Apr 15 '17 at 00:07
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    @imranfat Integral + comparison test is pretty Calc II in my opinion. Though this was a rather tricky one. – Simply Beautiful Art Apr 15 '17 at 00:09
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    @SimplyBeautifulArt Yes, the Comparison Test is standard, but I have never seen this kind of problem in calc2 textbooks. I hope a different (easier, yet mathematically correct) answer arises. Then my little mind can also grasp:) – imranfat Apr 15 '17 at 00:16
  • I can follow 80% of the answer given above. While I'm inclined to believe the answer it looks like a proof rather than a test. And our professor claims it can be done with what he has taught us in the calc II class. – Justin Clark Apr 15 '17 at 01:10
  • @SimplyBeautifulArt I'm wondering if this will work? If I set 0<|cos(ln(n))/n| <=1/n & sum from 1 to infinity of 1/n diverges by the p-test since p=1<=1 therefore, sum from 1 to infinity |cos(ln(n))/n| diverges by the direct comparison test. Thus sum from 1 to infinity cos(ln(n))/n diverges. – Justin Clark Apr 15 '17 at 01:30
  • Certainly the series doesn't converge absolutely, but I don't think this tells you anything about the series without the absolute values. Note that the sum of $\frac{\cos n}{n}$ (without the log) does converge. – Misha Lavrov Apr 15 '17 at 01:35
  • @JustinClark no, you have proven nothing. Recall that if $a_n<b_n$ and $\sum b_n$ diverges, then the comparison is inconclusive. – Simply Beautiful Art Apr 15 '17 at 10:36
  • While the above answer may be correct mathematically, the original question still isn't answered for a calc II level understanding of calculus. – Justin Clark Apr 16 '17 at 18:33