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.