Convergence of infinite series

[LoKI.GuZ]

Hi there,

I'm sorry if this is not be the best place to ask this:

Does the infinite series:

S_n = sum (n = 1 to infinity) cos (a * n) / n,

where a is a constant converge??

[jinydu]

I'm assuming you mean:

(cos (a*n))/n

This may take me a while to work on. I'll start off with the simplest possible case: taking a = 0. Thus, we have

S_n = sum (n = 1 to infinity) cos (0) / n

S_n = sum (n = 1 to infinity) 1/n

This is just the harmonic series. We already know it does not converge. Let me know if you would like a proof of this.

[LoKI.GuZ]

Originally Posted by jinydu

I'm assuming you mean:


(cos (a*n))/n)

Yeah, that's what I meant.. sorry for that...

My first idea was to use the integral test... but for that it would be necessary to calculate the cosine integral (as I just found in http://mathworld.wolfram.com/CosineIntegral.html)

Edit: It seems that the series converges for a = 1 (the integral does, to around 0.04)...

[mfgoode]

Originally Posted by jinydu

I'm assuming you mean:

(cos (a*n))/n

This may take me a while to work on. I'll start off with the simplest possible case: taking a = 0. Thus, we have

S_n = sum (n = 1 to infinity) cos (0) / n

S_n = sum (n = 1 to infinity) 1/n

This is just the harmonic series. We already know it does not converge. Let me know if you would like a proof of this.

What if a is not=0?
Mally

[Orgasmic Troll]

when a is of the form 2*k*pi, the series is the harmonic series, which doesn't converge, when a is of the form (2*k+1)*pi, the series is the alternating harmonic series, which converges to ln 2

I have a feeling that the series will converge for all a not of the form 2*k*pi, but that's just a hunch

[Orgasmic Troll]

here's a very slightly related topic:

http://www.stat.ualberta.ca/people/s...prints/rhs.pdf

its a paper on the random harmonic series, which converges most of the time, perhaps there is some way to expand this to replace the coin with a new "coin" that is a random variable distributed on [-1,1] and get similar results

then if you could show that for most a, cos(a*n) operates as a similar "coin" then you could show that it converges for most a

[mfgoode]

Originally Posted by TravisT

when a is of the form 2*k*pi, the series is the harmonic series, which doesn't converge, when a is of the form (2*k+1)*pi, the series is the alternating harmonic series, which converges to ln 2

I have a feeling that the series will converge for all a not of the form 2*k*pi, but that's just a hunch

There are flaws in the previous derivations.
To put the record straight,

When 'a' is an even multiple (2k) of pi we get the harmonic series and this is well known to diverge.

If 'a' is an odd multiple (2k + 1) of pi we get the negative harmonic series which is clearly seen to diverge negatively.

I 'a' is any integral 'k' multiple of pi we obtain the oscillating series.
This may be obtained by the expansion of log (1 + x) x < 1 and equal to log 2
so it converges.

Mally....

[jocelynl]

cos (a*n) is < 1 +-
So k/n when n is infinite is near 0
so yes it converge

[Orgasmic Troll]

Originally Posted by mfgoode

There are flaws in the previous derivations.
To put the record straight,

When 'a' is an even multiple (2k) of pi we get the harmonic series and this is well known to diverge.

If 'a' is an odd multiple (2k + 1) of pi we get the negative harmonic series which is clearly seen to diverge negatively.

I 'a' is any integral 'k' multiple of pi we obtain the oscillating series.
This may be obtained by the expansion of log (1 + x) x < 1 and equal to log 2
so it converges.

Mally....

Thank you, this is correct

[mfgoode]

Originally Posted by jocelynl

cos (a*n) is < 1 +-
So k/n when n is infinite is near 0
so yes it converge

You must bear in mind that we are dealing with a summation in the original expression and not a simlpe limit problem
Therefore
S(cos[an])/n is what we have to get
Now cos (an) =or< 1
Thfore cos(an)/n < 1/n
This is the harmonic series and diverges to infinity within the limits given
So all we can say in this method is that cos(an)/n < & (infinity).
It may or may not converge or diverge . We cannot say for sure.
In the method I have given above we can be sure.
Mally

[Zeta-Flux]

LoKI.GuZ,

The series does converge when a=1. This is a very hard result that one might prove in a course on Real Analysis.

I'd bet that one could modify the proof to work for some other a's.

Check out: http://at.yorku.ca/cgi-bin/bbqa?foru...0360.0001.0001

Best,
Zeta-Flux