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Tricks to Sum an Infinite Series


Date: 01/22/99 at 22:38:20
From: Steve Heggie
Subject: Convergence of a strange sum

   infinity
     SUM    [((-1)^n)*|sin n|]/n
    n = 1

Do you have any ideas on how to find the convergence of this series?  
I'm pretty sure it converges on zero, but I cannot begin to prove it 
(aside from just adding up the partial sums).

Thanks,
Steve


Date: 01/23/99 at 08:19:13
From: Doctor Anthony
Subject: Re: Convergence of a strange sum

The series is  -sin(1) + sin(2)/2 - sin(3)/3 + .......

    = -IM[e^(i) - e^(2i)/2 + e^(3i)/3 - ..... 

where IM means 'imaginary part of'. Let e^i = x and the series can be 
written:

    = -IM[x - x^2/2 + x^3/3 - x^4/4 + .......

Differentiating the series we get:

    = -IM[1 - x + x^2 - x^3 + .......

    = -IM[1/(1+x)]

To find the sum of the actual series we must integrate this, to get:

    = -IM[ln(1+x)]

    = -IM[ln(1+e^i)]

Suppose ln(1+e^i) = p+iq and we require the value of -q. Then:

    1 + e^i = e^(p+iq)

    1 + e^i = e^p e^(iq)

    1 + cos(1) + isin(1) = e^p[cos(q) + isin(q)]

    1 + cos(1) + isin(1) = e^p cos(q) + i e^p sin(q) 

Therefore:

    1 + cos(1) = e^p cos(q)
    sin(1) = e^p sin(q)

Dividing these equations:

    tan(q) = sin(1)/[1+cos(1)]

    q = tan^(-1)[sin(1)/(1+cos(1))] = 0.5

We require -q as the sum of the given series, so the required sum is 
-1/2.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 01/23/99 at 16:23:39
From: Steven Heggie
Subject: Re: Convergence of a strange sum

Dr. Anthony,

I've been working through the solution that you provided to my 
infinite sum problem ((-1)^n * |sin (n)|) / n, and I have figured 
everything out except for one part. It is the line that suggests that  
i sin(1) = i e^p sin(q). I don't understand why. Also, can you explain 
the line right after that?

Thank you,
Steve


Date: 01/23/99 at 17:49:55
From: Doctor Anthony
Subject: Re: Convergence of a strange sum

When you have a complex equation like:

    a + ib = p + iq

then we can equate the real part on the left with the real part on the 
right, that is, a = p, and equate the imaginary part on the left with 
the imaginary part on the right, that is, b = q.

It is easy to prove this. Write the equation:

    a-p = i(q-b)

Now square both sides:

    (a-p)^2 = -(q-b)^2    since i^2 = -1

    (a-p)^2 + (q-b)^2 = 0

This requires the sum of two positive terms to equal 0. Clearly this is 
impossible and in fact each bracket must separately equal 0. That is, 
a = p and b = q.

Applying this idea to the equation:

    1 + cos(1)+i sin(1) = e^p cos(q) + i e^p sin(q)

We must have:

    1 + cos(1) = e^p cos(q)    equating real part on each side
    sin(1) = e^p sin(q)        equating imaginary part on each side

I hope that has cleared up the difficulty.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 01/23/99 at 18:02:35
From: Steven Heggie
Subject: Re: Convergence of a strange sum

Dr. Anthony,

Thanks for the help. I NEVER would have figured this problem out on my 
own. Thank goodness for Euler's Formula!

- Steve Heggie
    
Associated Topics:
High School Sequences, Series

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