# Power Series

1. Oct 3, 2006

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

2. Oct 3, 2006

### StatusX

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.

3. Oct 3, 2006

(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?

4. Oct 3, 2006

### StatusX

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

5. Oct 5, 2006

So was I correct? Still not totally understanding it.

Thanks

6. Oct 5, 2006

### StatusX

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

7. Oct 6, 2006

### HallsofIvy

Staff Emeritus
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.