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Does the Series cos(n)/n^(3/4) Converge or Diverge?

Date: 11/10/2009 at 21:30:44
From: Collin
Subject: Is the series cos(n)/n^(3/4) convergent or divergent

Is the series cos(n)/n^(3/4) convergent or divergent?

This question is quite a doozy. I have tried the interval test and it 
was inconclusive.  I tried the strong comparison test but could not 
think of something to compare it to.  The limit comparison test gets 
too difficult as well.

Date: 11/12/2009 at 22:11:01
From: Doctor Jordan
Subject: Re: Is the series cos(n)/n^(3/4) convergent or divergent

Hi Collin,

The series cos(n)/n^(3/4) does, indeed, converge.  However, I will 
note that this is similar to the question of whether the series cos
(pi*n)/n^(3/4) is convergent.  That question is #15 in section 12.5 
of James Stewart's "Calculus," 6th ed.  For the latter series, we 
note that cos(pi*n) = (-1)^n, then find that the series is 
alternating; and since the terms decrease to 0, conclude that that 
series converges.

It is rather harder to show that cos(n)/n^(3/4) is convergent.  
Several times in the argument, we will have to bound expressions; for 
instance, 1 - cos(N + 1) <= 2.  I will explain how to show this; and 
if you want more details, please write back.

First we have to show that sum(cos(n), n = 0, ..., N) does not get 
too big.  This makes sense because cos(n) is bouncing around being 
negative and positive, so a sum of many cos(n) will probably not be 
that large, whether positive or negative.  It's not obvious at this 
point why we want to show that the sum of cos(n) from n = 0 to N does 
not get too big, but it will be useful later on in the proof.

We use the fact that e^(ix) = cos(x) + i*sin(x).  Then we use the 
finite geometric series:

    sum(x^n, n = 0, ..., N) = (1 - x^(N + 1))/(1 - x).

Here, we put x = e^i.  The real part of e^(i*n) is cos(n).  We take 
the real part of the above fraction, which will thus be equal to

    sum(cos(n), n = 0, ..., N).

After a few lines of work, we find that the real part of the fraction 

    (1 - cos(N + 1))*(1 - cos(1)) + sin(1)*sin(N + 1)
              (1 - cos(1))^2 + (sin(1))^2

So this is an expression for sum(cos(n), n = 0, ..., N).


    a_n = n^(-3/4)
    b_n = cos(n)
    B_N = sum(b_n, n = 1, ..., N)
    S_N = sum(a_n*b_n, n = 1, ..., N)
The complicated fraction with cos and sin that we found above is 
equal to B_N + cos(1), since B_N is summing from n = 1 to N instead 
of from n = 0 to N. Therefore,

                 (1 - cos(N + 1))*(1 - cos(1)) + sin(1)*sin(N + 1)
     B_N = -1 +  -------------------------------------------------
                            (1 - cos(1))^2 + (sin(1))^2

Let || denote absolute value.  We have to show that|S_N| has a limit 
as N -> infinity.  Now, to upper bound a fraction, we have to find an 
upper bound for the numerator and a lower bound for the denominator.

Since |cos(x)| <= 1 and |sin(x)| <= 1 for all x, using the triangle 
inequality we get the following:

                  2*2 + 1*1
    |B_N| <= 1 + -----------

Now, sin(1) > 4/5. So

    |B_N| < 1 + -----

          = 1 + 125/16

          = 141/16

          < 9. 

The formula for summation by parts is

    S_N = a_N*b_N - sum(B_n*(a_(n + 1) - a_n), n = 1, ..., N - 1).


    S_N <= a_N*b_N - 9*sum(a_(n + 1) - a_n, n = 1, ..., N - 1).

But all the terms in the sum cancel except for the first and last, 

    S_N <= a_N*b_N - 9*(a_N - a_1).

In other words,

    S_N <= N^(-3/4)*cos(N) - 9*N^(-3/4) + 9.

As N -> infinity, the right hand side approaches 9.  This shows that |
S_N| converges as N -> infinity. That is, the series cos(n)/n^(3/4) 
converges.  It doesn't show that its value is 9 -- only that its 
value is less than or equal to 9.

This was not an easy problem at all, so if you have any questions 
please feel free to write back.

- Doctor Jordan, The Math Forum 
Associated Topics:
High School Sequences, Series

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