# Interval of Convergence for a Series

• calculusisfun
In summary, the person is stuck at finding the interval of convergence for an inequality. They have reached the point of -4 < x^2 < 4 and are unsure how to proceed. Another person explains that the inequality can be solved by replacing the imaginary number with 0, giving the solution of -2<x<2. The person is grateful for the prompt response and understands the solution partially.

## Homework Statement

Ok, so I don't need help with this part, I just got stuck at the following step when attempting to find the interval of convergence:

## The Attempt at a Solution

I got here:

-4 < x^2 < 4

So, I need to solve this inequality. But can I? How can I take the square root of negative 4? And if this isn't possible to solve, what is the interval of convergence?

The inequality is perfectly easy to solve. It's -2<x<2. For every real number x^2 is greater than -4, so you don't even care about that limit.

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Thanks for the prompt response.

Okay, I understand what you're saying partially.

But to solve the inequality -4 < x^2 < 4, wouldn't you take the square root of both sides of the inequality to get the following:

root(-4) < x < root(4)

And root(-4) is an imaginary number is it not? Which would mean what for the interval of convergence?

calculusisfun said:
Thanks for the prompt response.

Okay, I understand what you're saying partially.

But to solve the inequality -4 < x^2 < 4, wouldn't you take the square root of both sides of the inequality to get the following:

root(-4) < x < root(4)

And root(-4) is an imaginary number is it not? Which would mean what for the interval of convergence?

Yes, sqrt(-4) is imaginary, but who cares? You aren't looking for imaginary solutions. You want real solutions. Just replace it with 0<=x^2<4.