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samtouchdown
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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.
sharks said: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?
There are two problems with this:sharks said: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.
sharks said: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?
I think i might have those two mixed up. What's the difference?Hurkyl said:You aren't talking about the convergence of the series: you're talking about the convergence of the sequence of terms
sharks said:I think i might have those two mixed up. What's the difference?
sharks said: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?
sharks said:So... The final verdict is? Converges or diverges?
Either way, I'm not sure i follow your method, Hurkyl.
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?Chirag B said: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.
lugita15 said: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?
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.Chirag B said: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?
The convergence of a series refers to the property that the sum of its terms approaches a certain finite value as the number of terms increases. On the other hand, the divergence of a series means that the sum of its terms does not approach a finite value as the number of terms increases.
To determine the convergence or divergence of a series, various mathematical tests can be used, such as the ratio test, comparison test, integral test, and limit comparison test. These tests evaluate the behavior of the series and can determine if it approaches a finite value or not.
Yes, a series can have both convergent and divergent parts. For example, the series 1+2+3+4+... is a divergent series, but if we remove every other term, we get the series 1+3+5+7+..., which is a convergent series. This shows that a series can have both convergent and divergent parts.
The sec(n)/n series is significant because it is an example of a divergent series. It is also known as the alternating harmonic series, and it is a counterexample to the harmonic series test for convergence. This means that just because the terms of a series are decreasing, it does not necessarily mean that the series is convergent.
In general, yes, the convergence or divergence of a series can change if we change the order of its terms. This is known as the reordering theorem. However, for some special series, such as absolutely convergent series, the convergence or divergence is not affected by the order of the terms.