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## 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?

• 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 < 3``3 > 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.
• 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.
• 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?
• what is the radius 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)
• You are just rearranging the function `h(x)`, to have it look as similar as possible to the formula of the sum of the geometric series. Since you are only rearranging, you cannot do anything that would change the value of the function, that is why you have to put the 2 minus signs, the first one is to make the function look like the formula, and the second one is to ensure that the value of the function doesn't change.