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### 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

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.

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

Sincerely,

Karthik
```
Associated Topics:
High School Sequences, Series

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