
#1
Mar3012, 11:56 PM

P: 10

I am wondering if this is solvable. Determine the convergene/divergence of the sum from n=1 to infinity of sec(n)/n. All the tests appear to fail and listing out the sequence of partial sums produces no useful results.




#2
Mar3112, 12:36 AM

PF Gold
P: 836

OK, you can rewrite your problem as [tex]\frac{1}{ncos(n)}[/tex] for starters.
Then as n tends towards infinity, you will get [itex]\frac{1}{∞}[/itex] which gives zero. Therefore the series converges. Maybe someone else can confirm or give you a better method? You could also try solving this using the Squeeze or Sandwich theorem: [tex]1 \leq \cos (n) \leq 1[/tex] Inverting gives: [tex]1 \leq \frac{1}{\cos (n)} \leq 1[/tex] Multiply by [itex]\frac{1}{n}[/itex] gives: [tex]\frac{1}{n} \leq \frac{1}{n\cos (n)} \leq \frac{1}{n}[/tex] Now, you just have to apply the theorem: For [itex]\frac{1}{n}[/itex], as n approaches infinity, the value tends towards 0. For [itex]\frac{1}{n}[/itex], as n approaches infinity, the value tends towards 0. Therefore, the middle term also gives the limit = 0, and the series converges. 



#3
Mar3112, 12:56 AM

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#4
Mar3112, 12:57 AM

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Convergence /Divergence of series:sec(n)/n




#5
Mar3112, 01:06 AM

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Let ε be small and positive. Then, if [itex]x \in (\frac{\pi}{2}  \epsilon, \frac{\pi}{2} + \epsilon)[/itex], we have [itex]\cos \epsilon < \epsilon[/itex]
(The intervals could be slightly bigger, but I doubt that extra precision is relevant) Since [itex]\cos x[/itex] is periodic with period π which is incommensurate with 1, we would expect that over a large interval of consecutive integer values of [itex]\cos x[/itex], the proportion of values less than [itex]\epsilon[/itex] should be at least [itex]2 \epsilon / \pi[/itex]. In particular, amongst the integers in [N, 2N) for large N, we would expect there to be roughly [tex]N \cdot \left( \frac{2 (1/N) }{\pi} \right) = \frac{2}{\pi}[/tex] points where [itex]\cos n < 1/N[/itex], and thus [itex]n \cos n < 2[/itex] So, it would be very surprising to find that [itex]n \cos n[/itex] converges as [itex]n \mapsto +\infty[/itex]. In fact, I honestly expect every real number to be a limit point. I'm pretty sure the holes in this proof can be sealed up; but it's been a long time since I've done a rigorous proof of this form so the method doesn't immediately spring to mind. Therefore, I'll leave it as an exercise. 



#6
Mar3112, 01:09 AM

PF Gold
P: 836

My apologies. I was just trying to help. I'm a student myself.
I suppose my 2nd suggestion of using the Sandwich theorem is also wrong, as well as the possible application of the comparison test? 



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Mar3112, 01:15 AM

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#8
Mar3112, 01:19 AM

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Mar3112, 01:21 AM

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#10
Mar3112, 01:45 AM

PF Gold
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OK, i understand. But how about using the nthterm test for divergence?
It would appear that the limit/sequence varies between 0 and infinity. Since the sequence does not go to zero, therefore the series diverges. This might prove useful for a more indepth solution: http://www.youtube.com/watch?v=yoldowfUGyQ Maybe the results can be confirmed if this problem is run through some math software? 



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Mar3112, 08:37 AM

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#12
Mar3112, 08:54 AM

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I can nearly prove the nterm test fails by invoking continued fractions: for any N, we can find a q such that
[tex]\left \pi  \frac{p}{q} \right < \frac{1}{q^2}[/tex] If p = 2m (and thus q is odd), we can multiply through by q/2: [tex]\left q \frac{\pi}{2}  m \right < \frac{1}{2q}[/tex] and get [tex]m \cos m < \frac{m}{2q} < 2[/tex] Convergents alternate between both p,q being odd, and exactly one of p,q being odd, but I wasn't able to rule out the strange possibility of q being the even one every time. (Edit: the above is wrong. (p,q) can go from (odd,even) to (even,odd) and vice versa. But the point is that p can't be even twice in a row, and the same for q) I imagine you could finish off the proof by using a suitable lattice instead of invoking continued fractions, or maybe there's a trick with continued fractions I missed to show p has to be even infinitely often. I thought about interpolating and using (p+p')/(q+q') for successive convergents p/q and p'/q', but I couldn't rule out the possibility that q' was much larger than q and spoiling the inequalities. This does leave me curious about the possibility of faster growing coefficients such as [itex]n^2 \cos n[/itex] could actually converge to [itex]+\infty[/itex]; i.e. that [itex]n^2[/itex] races off to infinity faster than n can approximate [itex]\pi/2[/itex] (modulo [itex]\pi[/itex]). 



#13
Mar3112, 09:03 AM

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Of course, as soon as I write this, I see what I was missing in my earlier attempts to wrap this up. If I approximate instead
[tex]\left \frac{\pi}{2}  \frac{p}{q} \right < \frac{1}{q^2}[/tex] then I can pick arbitrarily large convergents where q is odd. Then, [tex]\left q \frac{\pi}{2}  p \right < \frac{1}{q}[/tex] and [tex]p \cos p < \frac{p}{q} < 2[/tex] 



#14
Mar3112, 09:55 AM

P: 38

Just to point out why the squeeze theorem doesn't work, if
[tex] 0<x<y [/tex] [tex] x \frac{1}{xy} < y \frac{1}{xy} [/tex] [tex] \frac{1}{y} < \frac{1}{x}[/tex] Applied to [tex] 1 ≤ cos(n) ≤ 1 [/tex] [tex] \frac{1}{cos(n)} ≥ 1[/tex] and[tex] \frac{1}{cos(n)} ≤ 1 [/tex] 



#15
Mar3112, 10:13 AM

PF Gold
P: 836

So... The final verdict is? Converges or diverges?
Either way, i'm not sure i follow your method, Hurkyl. 



#16
Mar3112, 11:07 AM

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The plan is to use continued fractions to find a good approximation to [itex]\pi / 2[/itex], use that to find an integer for which [itex]\cos n[/itex] is small, and then observe that [itex]n[/itex] is not too big, so that [itex]n \cos n < 2[/itex]. 



#17
Mar3112, 08:34 PM

P: 18

Samtouchdown, I would try something along these lines:
We want to determine the convergence/divergence of the series [itex]\sum^{∞}_{n=1}\frac{sec(n)}{n}[/itex]. You said writing out terms is of no help. Perhaps it is not? Let's try first. [itex]\sum^{∞}_{n=1}\frac{sec(n)}{n} = sec(1) + \frac{sec(2)}{2} + \frac{sec(3)}{3} + \cdots[/itex]. If we carry this out to infinity we will see that eventually, the series will have to diverge, as the terms will collectively approach a number without a bound. Therefore, the series will diverge. However, it is possible that I may be wrong. In such a case, I would use the formal definition of a series to find the correct answer (the one that resembles the deltaepsilon definition of a limit). 



#18
Mar3112, 10:49 PM

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