What can be said about the convergence or divergence of the following series?

[courtrigrad]

Suppose that \sum_{n=0}^{\infty} c_{n}x^{n} converges when x=-4 and diverges when x=6. What can be said about the convergence or divergence of the following series?

(a) \sum_{n=0}^{\infty} c_{n}

(b) \sum_{n=0}^{\infty} c_{n}8^{n}

(c) \sum_{n=0}^{\infty} c_{n}(-3)^{n}

(d) \sum_{n=0}^{\infty} (-1)^{n}c_{n}9^{n}So we know that \sum_{n=0}^{\infty} c_{n}x^{n} converges when -5\leq x\leq5, and diverges when x> 5.

(a) Would \sum_{n=0}^{\infty} c_{n} diverge?
(b) This would diverge because x>5?
(c) This would converge, because -5<-3<5?
(d) This would diverge because x>5?

Thanks

 

[StatusX]

So we know that \sum_{n=0}^{\infty} c_{n}x^{n} converges when -5\leq x\leq5, and diverges when x> 5.

This doesn't follow from what was given. To show the convergent series converge, all you need to know is that |c_n (-4)^n| \rightarrow 0 as n \rightarrow \infty. For the divergent ones, show that if a certain series a_n diverges, then so does the series r^n a_n whenever |r|>1.

 

[courtrigrad]

(a) \sum_{n=0}^{\infty} c_{n} diverges because it \rightarrow \infty?

(b) \sum_{n=0}^{\infty} c_{n}8^{n}. How would I use this |c_n (-4)^n| \rightarrow 0 to establish that it converges?

For the rest, they arent geometric series, right?

 

[StatusX]

You have those backwards. Remember it's 4^n, not 1/4^n.

 

[courtrigrad]

So was I correct? Still not totally understanding it.

Thanks

 

[StatusX]

No. c_n 8^n grows faster than c_n 6^n, so the former diverges if the latter does.

 

[HallsofIvy]

If you are working with power series, you should have, long ago recognized that integers are not the only numbers that exist!

Knowing that the series does not converge for x= 6 does NOT mean that it only converges for x\le 5! It might converge for every x< 6.