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## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 14: Function as a geometric series

# Geometric series interval of convergence

When a power series is a geometric series, we can find its interval of convergence without using the ratio test! Created by Sal Khan.

## Want to join the conversation?

• don't you need to check the limits of the interval to see whether or not it converges to the point?
• actually, he didnt check the endpoints of the interval because this is a geometric series. since |r|<1 and not < or = to 1, the endpoints will cause the series to diverge because they are not included.
• At , how is the abs(-x^2/3) the same as abs(x^2/3) ? Wouldn't the negative sign have to be like this: abs{(-x^2)/3}, for the negative sign to disappear since -x times -x is positive x^2? I know this is very basic, but it's confusing me...
• Let's solve it in the normal way:
|-x²/3| < 1-1 < -x²/3 < 1-3 < -x² < 33 > x² > -3

Because x² >= 0 > -3 for every x so we just need to take care x² < 3-√3 < x < √3.
• Someone help me understand:
So the point of this video is to demonstrate how if we take a function, and then manipulate it in a way that it looks like a/1-x, and if we take the infinite sum of that function, then sum formula will be the same as the original function we had in the beginning?
• Yes - sometimes it's best to solve a problem by manipulating the function to look like a well-known model function that is easy to work with, such as a/1-r. The original function and the new one are equivalent. This technique is also used, for example, to integrate inverse trig functions such as arctan x.
• Can't we also define the function as 1 / [ 1- (-2 - x^2)] ? In which case the interval of convergence would be |2 + x^2| < 1. How do you reduce this? This doesn't seem correct cuz 2 + x^2 is always positive and 2 + x^2 < 1 which implies x^2 < -1 ! So this first representation is wrong because I should not have any constant terms in the common ratio? Can someone please explain this?
• Not all series converge everywhere: your series happens to converge nowhere (in the real numbers, that is.). That's all there is to it: not everything necessarily works out.
(1 vote)
• Is it correct to say that for functions such as the ones shown in the video the interval of convergence is the domain of the function ? Thanks for the patience :)
• Let's see.
f(x) = 1 / (1 - x)
That converges for -1 < x < 1.
Is f(x) defined for |x| > 1 ? It seems to work for any x except 1, so I'd say yes. Hence, the domain is not restricted to the interval of convergence.
• If x = 0, then the sum's first term will be (1/3)*0^0 power. 0^0 is undefined, so shouldn't we exclude that from the interval of convergence?
• I should understand this, but at how does Sal get [1/3] / [ 1 + x ] ?? More specifically I don't understand how he got the 1/3 on top. Any help would be great!
• consider a simpler situation: 1 divided by 12.
1/ (3.4)
= 1/3 (4) = (1/3) /4 ( same as 1/3 divided by 4)
• what is the radius of convergence?