Thread: Why does the sequence of a power series converge in the following situation?

1. Why does the sequence of a power series converge in the following situation?

So the power series $\sum_{n=0}^{\infty}a_n (x-4)^n$ converges at 0 and diverges at 9. How do we know that the limit of the sequence $a_n$ is zero? Thanks.

2. Originally Posted by tsp216
So the power series $\sum_{n=0}^{\infty}a_n (x-4)^n$ converges at 0 and diverges at 9. How do we know that the limit of the sequence $a_n$ is zero? Thanks.
"converges at 0" What do the terms look like for x = 0?

3. Originally Posted by tkhunny
"converges at 0" What do the terms look like for x = 0?
@tkhunny so I guess that would be
$a_0+a_1(-4)+a_2(16)+a_3(-64)...$
I still don't see how the limit of the sequence $a_n$, $\lim\limits_{n\to\infty}a_n$ is 0?

4. Originally Posted by tkhunny
"converges at 0" What do the terms look like for x = 0?
So the sequence would look like
$a_0+a_1(-4)+a_2(16)+a_3(-64)+...$

I still don't see how the limit of $a_n$, $\lim_\limits{n\to\infty}a_n$, is 0?
EDIT: @ksdhart2 I've made the correction to the typo.

5. Originally Posted by tsp216
So the sequence would look like
$a_0+a_1(-4)+a_2(16)+a_3(-64)+...$

I still don't see how the limit of $a_n$, $\lim_\limits{x\to\infty}a_n$, is 0?
Well, first off I think there was a typo there and you meant to say $\displaystyle \lim_{n \to \infty} a_n = 0$. Assuming that's the case, let's begin by reformulating the terms slightly. Let $b_n = a_n \cdot (-4)^n$ such that the series becomes:

$b_0 + b_1 + b_2 + b_3 + b_4 + ...$

Now we've already established that this series converges, so what must be true about $\displaystyle \lim_{n \to \infty} b_n$? Recall back to when you were first learning about series and you learned all those convergence tests. In particular, I'm thinking of the Divergence Test. From the results of this test, what can you then say about $\displaystyle \lim_{n \to \infty} a_n$?

6. My understanding is that the divergence test only tells you that if the limit of the sequence $a_n$ is not zero, then the series diverges, and that it doesn't tell us that given a series that converges, its sequence must equal 0, which means that knowing that the series converges does not tell me that the sequence $a_n$ converges, is this correct?

7. Originally Posted by tsp216
@tkhunny so I guess that would be
$a_0+a_1(-4)+a_2(16)+a_3(-64)...$
I still don't see how the limit of the sequence $a_n$, $\lim\limits_{n\to\infty}a_n$ is 0?
That's right. So, if the whole thing converges, what do the $a_{n}$ have to do?

8. Actually, having re-read that link to Paul's Online Maths Notes, it is indeed that a convergent series has a sequence that tends to 0. I can see the solution to the original problem now. However, isn't the power series, rather than the sequence $a_n$, that converges, and so wouldn't the theory only apply to the power series, and so we still can't conclude anything about $a_n$ itself?

9. Originally Posted by tsp216
Actually, having re-read that link to Paul's Online Maths Notes, it is indeed that a convergent series has a sequence that tends to 0. I can see the solution to the original problem now. However, isn't the power series, rather than the sequence $a_n$, that converges, and so wouldn't the theory only apply to the power series, and so we still can't conclude anything about $a_n$ itself?
If we take your original words at face value, "converges at zero", we can say something of the $a_{n}$, and we just did.

The first requirement of a convergent series is the convergence of the individual terms to zero. If that doesn't happen, the sequence of partial sums cannot converge.

More specifically, in this case, we know the series converges at x = 0. The sequence of terms has two pieces 1) $a_{n}$ and 2) increasing multiples of 4. Thus $a_{n} \cdot 4^{n}$ MUST find its way to zero (0). We know $4^{n}$ doesn't do that. What can we conclude about the $a_{n}$?

10. @tkhunny great, I've got it now, thanks for clearing that up. It feels good to get that annoying bit of confusion out of the way.