# algebra

posted by
**Jane E.** on
.

Okay, I was taught that if a geometric series is infinite and it diverges then it has no limit and no sum. Looking back in my notes, I found an example of finding the value for a divergent series? Is this possible?

The sum of an infinite geometric progression is infinite when the common ratio, r, is greater than 1.

The sum of an infinite geometric progression is finite and given by Sn = a/(1-r) when -1 < r < +1.

Yes, divergent series can sometimes be "resummed" to yield a finite value. Resumation techniques are used quite often in physics. E.g., suppose you want to calculate the pressure of a gas as a function of volume and temperature. The interactions between the gas molecules make this problem impossible to solve exactly. However, you can solve the problem exactly if the interaction is zero. Using some tricks you can in fact derive a series in powers of the parameter

g = (interaction strength)/(temperature).

But this series does not converge if g is large, e.g. when the temperature becomes low. However for large g the pressure will still be some function of g. One can find this function by making certain assumptions about how the function should behave for large g (e.g. you know that it becomes a fluid).

Then you can find a suitable function which has the correct series expansion for small g. That function can then be used to calculate the pressure in the fluid phase.

In case of a geometric series:

f(x)= 1 + x + x^2 + ...

the function f(x) is given by:

f(x) = 1/(1-x)

f(x) is still defined for |x|>1 even though the series does not converge there.

1/(1-x) can be called the resummed value of the series 1 + x + x^2 + ... for |x|>1

okay thanks