Question

Will award medal
the power series 􏰂 Cnxn diverges at x = 7 and converges at x = −3. At x = −9, the series is
Conditionally convergent Absolutely convergent Divergent
Cannot be determined.
@ ganeshie

Answer 1

I know there is a visual way to represent this but i just do not remember. something with a number line and radius

Answer 2

@Zarkon

Answer 3

@SithsAndGiggles

Answer 4

You're dealing with the given series' interval of convergence.

By the ratio test, you have
\[\lim_{n\to\infty}\left|\frac{c_{n+1}x^{n+1}}{c_nx^n}\right|=|x|L\]
where \(L\) is the limit of the sequence \(\dfrac{c_{n+1}}{c_n}\). The series is known to converge for \(|x|L<1\), or \(|x|<\dfrac{1}{L}\), or \(-\dfrac{1}{L}<x<\dfrac{1}{L}\) (this is the convergence condition for the ratio test).

Answer 5

Knowing that the series diverges when \(x=7\), you then know that \(|7|=7\not<\dfrac{1}{L}\), so that you must have \(L\ge\dfrac{1}{7}\).

Knowing that the series converges at \(x=-3\), you have \(|-3|=3<\dfrac{1}{L}\), so \(L<\dfrac{1}{3}\).

So: You know that \(L\) is bounded, with \(\dfrac{1}{7}\le L< \dfrac{1}{3}\). Remember, you have to satisfy \(|x|L<1\) in order for the series to converge. What happens when \(x=-9\)?