Arithmetico-Geometric Series and Polylogarithms

Date: 07/06/2006 at 16:27:43
From: Karthik
Subject: What is the sum of the following infinite series

In my research about microprocessor temperature behaviour, I 
encountered a series similar to the one below:

  e^(-x) + 1/9 * e^(-3x) + 1/25 * e^(-5x) + 1/49 * e^(-7x) + ...

Is there a closed form expression for the sum of this series?  I find
the combination of the geometric series e^-(2n+1)x and the series
1/(2n+1)^2 to be the most confusing part.

1) Using Fourier series expansion of a saw-tooth waveform, I could 
derive the following relation:

  1 + 1/9 + 1/25 + 1/49 + ... = pi^2 / 8

2) The given series can be bounded from above by the geometric series 
(as every individual term is larger than that of the given series): 
e^(-x) + e^(-3x) + e^(-5x) + ... .  Using the formula for the infinite 
sum of a geometric series, this implies that the series sum is less 
than or equal to e^(-x) / (1 - e^(-2x)).

3) Similarly, the series can be bounded from below by the geometric 
series e^(-x)/e + e^(-3x)/e^3 + e^(-5x)/e^5 + ... as every term in 
it is smaller than that of the given series.  Hence, the series sum 
is greater than or equal to e^-(x+1) / (1 - e^-2(x+1)).

Thus, I am able to bind the series sum between f(x) and f(x+1) where 
f(x) = e^(-x) / (1-e^(-2x)).  However, I am not able to close the gap 
further and find an exact expression for the sum.  Could you please 
help me with it?  Thank you.

Date: 07/07/2006 at 18:21:55
From: Doctor Vogler
Subject: Re: What is the sum of the following infinite series

Hi Karthik,

Thanks for writing to Dr. Math.  The answer to your question is "sort 
of, but not really."

If the 9, 25, 49... were in the numerators instead of the 
denominators, then you would have an arithmetico-geometric series 
using

  t = e^(-x)

and could sum it using the technique of

  Finding the Sum of Arithmetico-Geometric Series
    http://mathforum.org/library/drmath/view/66996.html 

As it is, this is a good start.  In any case, here is a method that 
can be very helpful in solving problems like yours, although it has a 
snag in your case.

Set t = e^(-x) as above and consider the function

  f(t) = t + (1/3^2)t^3 + (1/5^2)t^5 + (1/7^2)t^7 + ....

Notice that this function is the sum you want.  So we just need to get 
a closed-form expression for f(t).  Well, if we could get rid of the 
coefficients, then we would have a geometric series.  So how do we get 
rid of the coefficients?  Answer:  To multiply by the exponent of t, 
we take a derivative:

  f'(t) = 1 + (1/3)t^2 + (1/5)t^4 + (1/7)t^6 + ....

Wow!  That was amazing!  Now we just need to do it one more time, 
except that the exponents are wrong now.  But they are all off by one, 
and there is an easy way to increase the exponent of t by one: 
Multiply by t.

  f'(t)*t = t + (1/3)t^3 + (1/5)t^5 + (1/7)t^7 + ....

We're almost there!  Now we take another derivative, and we have a 
geometric series:

  (f'(t)*t)' = 1 + t^2 + t^4 + t^6 + ...
             = 1/(1 - t^2)

Finally, we have a closed-form expression!  Now we just need to 
reverse our operations on the closed-form expression in order to 
compute f(t).  So first we take an antiderivative:

  f'(t)*t = integral of 1/(1 - t^2) dt
          = integral of (1/2)( 1/(1+t) + 1/(1-t) ) dt

  f'(t)*t = (1/2)(ln(1+t) - ln(1-t)) + C.

Looking back at our series for

  f'(t)*t = t + (1/3)t^3 + (1/5)t^5 + (1/7)t^7 + ...,

we see that we want the right side to evaluate to 0 when t=0, which 
means that we should take C = 0.  So we have

  f'(t)*t = (1/2)(ln(1+t) - ln(1-t)).

Solving for f'(t) is easy:

  f'(t) = (1/2)(1/t)(ln(1+t) - ln(1-t)).

Now, we need to take one more antiderivative, and then we'll be done!  
Unfortunately, the antiderivative can't be expressed in terms of 
"elementary" functions.  You could do it using polylogarithms

  Polylogarithms
    http://mathworld.wolfram.com/Polylogarithm.html 

but I can't think of why that's really any better than the infinite 
series that you started with.

In any case, I hope that I have at least taught you a few new 
techniques for solving series, even if we couldn't put yours in a 
nice form.

If you have any questions about this or need more help, please write 
back and show me what you have been able to do, and I will try to 
offer further suggestions.

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

Date: 07/07/2006 at 20:17:01
From: Karthik
Subject: Thank you (What is the sum of the following infinite series)

Dear Doctor Vogler,

Thank you very much for your help.  It taught me the interesting
technique of solving arithmetico-geometric series.  Thank you!

My roommate and I tried the derivative technique but got stuck in the
integral process (not being aware of polylogarithms).  I will work on
this more and contact you again if I have any questions.  Thanks again
for your help. 

Sincerely,

Karthik