# Homework Help: Convergence /Divergence of series:sec(n)/n

1. Mar 30, 2012

### samtouchdown

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. Mar 31, 2012

### sharks

OK, you can rewrite your problem as $$\frac{1}{ncos(n)}$$ for starters.

Then as n tends towards infinity, you will get $\frac{1}{∞}$ 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:

$$-1 \leq \cos (n) \leq 1$$
Inverting gives:
$$-1 \leq \frac{1}{\cos (n)} \leq 1$$
Multiply by $\frac{1}{n}$ gives:
$$-\frac{1}{n} \leq \frac{1}{n\cos (n)} \leq \frac{1}{n}$$
Now, you just have to apply the theorem:
For $-\frac{1}{n}$, as n approaches infinity, the value tends towards 0.
For $\frac{1}{n}$, as n approaches infinity, the value tends towards 0.
Therefore, the middle term also gives the limit = 0, and the series converges.

Last edited: Mar 31, 2012
3. Mar 31, 2012

### Dick

That does not work. On several levels. The first is that the sequence converging to zero doesn't show the sum of the series converges. And for another thing it's not at all clear n*cos(n) goes to infinity. I don't know how to do this one. But if you could prove that the sequence sec(n)/n does not go to zero, that would prove the series doesn't converge. I suspect this is true.

4. Mar 31, 2012

### Hurkyl

Staff Emeritus
There are two problems with this:
• You aren't talking about the convergence of the series: you're talking about the convergence of the sequence of terms
• $\lim_{x \rightarrow +\infty} x \cos x$ doesn't exist
(I'm not sure if the limit exists if x is restricted to integers, but I would be mildly surprised if it did)

5. Mar 31, 2012

### Hurkyl

Staff Emeritus
Let ε be small and positive. Then, if $x \in (\frac{\pi}{2} - \epsilon, \frac{\pi}{2} + \epsilon)$, we have $|\cos \epsilon| < \epsilon$

(The intervals could be slightly bigger, but I doubt that extra precision is relevant)

Since $|\cos x|$ is periodic with period π which is incommensurate with 1, we would expect that over a large interval of consecutive integer values of $|\cos x|$, the proportion of values less than $\epsilon$ should be at least $2 \epsilon / \pi$.

In particular, amongst the integers in [N, 2N) for large N, we would expect there to be roughly
$$N \cdot \left( \frac{2 (1/N) }{\pi} \right) = \frac{2}{\pi}$$
points where $|\cos n| < 1/N$, and thus $|n \cos n| < 2$

So, it would be very surprising to find that $n \cos n$ converges as $n \mapsto +\infty$. 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. Mar 31, 2012

### sharks

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?

7. Mar 31, 2012

### Dick

Sure it is. cos(x) can be very small number. So 1/cos(x) can be a very large number. That's how -1<=1/cos(x)<=1 can be wrong. You don't really need to apologize.

8. Mar 31, 2012

### sharks

I think i might have those two mixed up. What's the difference?

9. Mar 31, 2012

### Dick

1/n converges to zero as a sequence, the sum of 1/n diverges because it's a harmonic series.

10. Mar 31, 2012

### sharks

OK, i understand. But how about using the nth-term 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 in-depth solution:

Maybe the results can be confirmed if this problem is run through some math software?

Last edited by a moderator: Sep 25, 2014
11. Mar 31, 2012

### Dick

Yeah, the nth term test is what I was suggesting in post 3. I'm pretty sure it does work. But I don't know how to prove it.

Last edited by a moderator: Sep 25, 2014
12. Mar 31, 2012

### Hurkyl

Staff Emeritus
I can nearly prove the n-term test fails by invoking continued fractions: for any N, we can find a q such that

$$\left| \pi - \frac{p}{q} \right| < \frac{1}{q^2}$$

If p = 2m (and thus q is odd), we can multiply through by q/2:

$$\left| q \frac{\pi}{2} - m \right| < \frac{1}{2q}$$

and get

$$|m \cos m| < \frac{m}{2q} < 2$$

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 $|n^2 \cos n|$ could actually converge to $+\infty$; i.e. that $n^2$ races off to infinity faster than n can approximate $\pi/2$ (modulo $\pi$).

Last edited: Mar 31, 2012
13. Mar 31, 2012

### Hurkyl

Staff Emeritus
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

$$\left| \frac{\pi}{2} - \frac{p}{q} \right| < \frac{1}{q^2}$$

then I can pick arbitrarily large convergents where q is odd. Then,

$$\left| q \frac{\pi}{2} - p \right| < \frac{1}{q}$$

and

$$|p \cos p| < \frac{p}{q} < 2$$

14. Mar 31, 2012

### Bearded Man

Just to point out why the squeeze theorem doesn't work, if
$$0<x<y$$
$$x \frac{1}{xy} < y \frac{1}{xy}$$
$$\frac{1}{y} < \frac{1}{x}$$

Applied to
$$-1 ≤ cos(n) ≤ 1$$
$$\frac{1}{cos(n)} ≥ 1$$
and$$\frac{1}{cos(n)} ≤ -1$$

Last edited: Mar 31, 2012
15. Mar 31, 2012

### sharks

So... The final verdict is? Converges or diverges?

16. Mar 31, 2012

### Hurkyl

Staff Emeritus
The basic ideas are:

• Stepping along 1 unit at a time, you can get arbitrarily close to any point in the period of $|\cos x|$.
• If $|x - \pi/2| < \epsilon$, then $|\cos x| < \epsilon$
• Continued fractions are a standard technique to produce very good rational number approximations to numbers

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

17. Mar 31, 2012

### Chirag B

Samtouchdown, I would try something along these lines:

We want to determine the convergence/divergence of the series $\sum^{∞}_{n=1}\frac{sec(n)}{n}$.

You said writing out terms is of no help. Perhaps it is not? Let's try first.

$\sum^{∞}_{n=1}\frac{sec(n)}{n} = sec(1) + \frac{sec(2)}{2} + \frac{sec(3)}{3} + \cdots$.

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 delta-epsilon definition of a limit).

18. Mar 31, 2012

### lugita15

But you're asserting the very thing we need to prove. How do you know that the sequence of partial sums goes to infinity without applying a convergence test?

19. Apr 1, 2012

### Chirag B

That's a good point. It appears I was mistaken. But all convergence/divergence tests appear to fail here. Is there perhaps another way to do it?

20. Apr 1, 2012

### lugita15

The nth term test may succeed here, although it's a bit hard to find the limit of the nth term, as Hurkyl is trying to do.