Does this series converge or diverge?

lillybeans

1. The problem statement, all variables and given/known data



3. The attempt at a solution

I really couldn't figure this one out. It's definitely not telescoping series because I don't see repeating terms that cancel. I would try to compare it to a geometric series. so what I did was factoring out a 4^n:

4^n(2+sinn-(1/2+sinn))

replacing (2+sinn-(1/2+sinn)) with "a"

Then I noticed the bounds 1<2+sinn<3

so the whole thing inside the bracket is oscillating between 0<a<8/3

So i concluded that the series is divergent? (highly unsure) because the coefficient that 4^n multiplies by (by coefficient I mean "a") is always either 0 or a positive number. So when it is 0, there will be no effect. When it is a positive number, the sums will add up, and the series will eventually go to infinity.. Eh.. is that.. a valid line of reasoning?

Thanks in advance,

Lilly.

Bohrok

Since 4ⁿ→∞ as n→∞, the (2+sin n) - 1/(2+sin n) part had at least better go to 0. Can you show whether the limit of that part is 0 as n→∞?

lillybeans

I am quite sure the limit of the bracket part doesn't go to zero because sin(n) is always oscillating.

But is it supposed to?

Bohrok

L'Hopital's rule doesn't apply since we don't have either of the indeterminate forms 0/0 or ∞/∞ (usually we can't use it when we have trig functions in the fraction like that), so we'll have to try something else.

2+sin n is definitely oscillating, but we can't really assume either way that the -1/(2+sin n) would make it tend to 0 or oscillate still.
Still, if you look at a graph, it looks just like a sine or cosine graph that's been translated and stretched, but I have no idea how you could write it as just one trig function...

Bohrok

You can show that the derivative of (2+sin x) - 1/(2+sin x) oscillates, so (2+sin n) - 1/(2+sin n) doesn't tend to 0, and so the series diverges.

lillybeans

Got it, thank you very much

Office_Shredder

The derivative of a function does not have to go to zero for the function itself to go to zero

Also, It's not enough to show that the continuous version of the function oscillates. For example [itex] \lim_{x\to \infty} cos(2 \pi x)[/itex] does not exist, but [itex] \lim _{n \to \infty} cos(2\pi n) = 1[/itex].

The argument to show that 2+sin(n)-1/(2+sin(n)) does not go to zero is only slightly non-trivial. Suppose that this is close to zero. Then 2+sin(n) is approximately 1/(2+sin(n)) so it must be that 2+sin(n) is approximately 1 and sin(n) is approximately -1. Hence it must be that n is very close to 3pi/2+2kpi for some k. Based on this prove that 2+sin(n+1)-1/(2+sin(n+1)) cannot be close to zero

lillybeans

I see. Thanks for the reply.

One additional question: Can you solve this question by using any of the following techniques? These are the techniques we learned in class, so it is most probable that our professor wants to test us on one of them, but it could be an anomaly.

- Comparison with geometric
- Comparison with p-series
- Ratio test
- Root test
- Divergence test
- Alternating series
- Limit comparison

Bohrok

O_S, I see that we can't just assume as you showed with that example.
The derivative in this case turned out to be (1+1/(sinx+2)²)cosx. The part in front is bounded above by 2 and bounded below by 10/9, so what else could be done to show that this doesn't settle on any number?