Convergence Test for Series with Cosine Cubed Terms

[peripatein]

Hi,
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibniz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)

 

[Mark44]

Hi,
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibniz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)

The series is not alternating, despite the factor of (-1)ⁿ, so I don't see that the Leibniz Criterion applies. I also don't see that the hint is helpful.

For any n, the terms in your series are bounded below by -1/(3ⁿ) and bounded above by 1/(3ⁿ), both of which are terms in geometric series.

 

[SammyS]

Hi,
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibniz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)

Is that really an x in the series?

$\displaystyle (-1)^n\frac{\cos^3( (3^n)x)}{3^n}$​

If so, the divergence/convergence certainly may depend upon x.

 

[peripatein]

That is indeed x in the formula.
Mark44, I am still not sure how to go about it. May you please assist and/or elaborate further?

 

[haruspex]

The hint is very useful. You have a cos³() in the expression for the series, and the hint gives you a substitution for that. Try it.

 

[SammyS]

As Mark said, the series is not an alternating series.

In fact it is alternating for a few specific values of x, namely when x is an integer multiple of π .

It can also be a constant series with 0 as the constant if x is an odd half-integer multiple of π .

... But to help you carry on with the problem:

What is the range of cos(x), and cos³(x) ?

Use the squeeze theorem.

 

[peripatein]

Why would the sum of this series be equal to 0.75cos(x)? I managed to express this series as a telescopic summation and thus merely (1/4)[-cos3x - ((1/3)^n)cos(3^(n+1)x)] remain in the sum. Won't ((1/3)^n)cos(3^(n+1)x) be equal to zero as n->infinity? Hence, why won't the sum be -0.25cos3x?

 

[haruspex]

I managed to express this series as a telescopic summation and thus merely (1/4)[-cos3x - ((1/3)^n)cos(3^(n+1)x)] remain in the sum.

Does the sum start at n=0 or n=1? That would account for the different results.