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### Course: AP®︎/College Calculus BC > Unit 10

Lesson 13: Radius and interval of convergence of power series# Worked example: interval of convergence

The interval of converges of a power series is the interval of input values for which the series converges. To find it, we employ various techniques. See how it's done in this video.

## Want to join the conversation?

- How would I find the radius of convergence for this series?(7 votes)
- The radius of convergence is half of the interval of convergence. In the video, the interval is -5 to 5, which is an interval of 10, so the radius of convergence is 5.

(This is unaffected by whether the endpoints of the interval are included or not)(20 votes)

- Starting from6:16, i don't understand about the harmonic series and p-series. Why in p-series when p is equal to 1, it must be diverging ? thanks.(12 votes)
- When p = 1, it is just 1/n, right? if you add them up all the way to infinity, that IS just the famous harmonic series! Here is the link that Sal proves that harmonic series indeed, diverges! It is counter-intuitive that it diverges, but hopefully this video satisfies you ^^);

https://www.khanacademy.org/math/calculus-home/series-calc/convergence-divergence-tests-calc/v/harmonic-series-divergent(10 votes)

- shouldn't we test the end points? or do we always not include them?(7 votes)
- Yes, the end points need to be individually tested. Sometimes they're included, sometimes they're not, sometimes only one is and the other is not.(9 votes)

- This might be a stupid question but are there any function series whose convergence intervals are not contiguous? If no, why?(7 votes)
- It's a good question. I think the series Simga(n = 1->infinity) x^(-n) would have an interval of convergence: (-inf, -1) U ( 1, inf) since the common ratio of the series would be 1/x

(i know this is really late but I'm posting for reference in case it helps anyone in future)(2 votes)

- how does 5^n/5^n+1 = 1/5(3 votes)
- great question! Think of it this way (sorry for the weird formatting):

When n=1, 5^n is just 5, right?

What about 5^(n+1)? If n=1, wouldn't that be 5^2 = 25?

Recall this rule of exponents: (X)^a•(X)^b = (X)^(a+b)

Therefore: (5)^n•(5)^1 = (5)^(n+1).

So, if you reverse that rule, it works this way:

while n=1, 5^(n+1) = 5•5^(n)

If you had a larger number, say 5^(n+7), then it devolves into 5•5^(n+6), because 5^1 = 5.

In the end, (5^n)/(5^[n+1]) = (5^n)/(5•5^n).

Hope this helps! Best of luck!(7 votes)

- When i do the ratio test and end up with an answer that doesnt include n does that mean that it diverges and has no interval of convergence? Or does that mean that it converges over any interval to that answer? For example
`2^n*x^2n`

or`(x^n)/(2^n)`

(4 votes)- If the answer is smaller than 1, the series always converges; if the answer is greater than 1, the series always diverges. It is inconclusive if the answer is equal to 1.

In other words, if it doesn't include n you still use the Ratio Test as usual. The limit as n approaches infinity won't change anything.(1 vote)

- 7:55Monotonically?(1 vote)
- Monotonically decreasing means strictly decreasing; there's no point where the n+1th term is greater than the nth term.(4 votes)

- If my limit evaluates to 0 does that mean that it converges over any interval?(2 votes)
- My answer is going to be yes, but I also want to say "for some cases" because I haven't seen enough cases. Take for example the infinite series of (x^k)/k!. You'll notice that |x| times the limit as k approaches infinity is equal to 0, meaning for any value of x you pick, you'll always find that r=0 for all real numbers, so the IOC would be over all reals.(2 votes)

- Does the interval of convergence always have to be symmetrical about c? That is, do the endpoints of the interval of convergence always have to be equidistant from c?(2 votes)
- No. I have had a problem before where I was given the task to find the interval of convergence for a Taylor series centered at 1, but the interval of convergence 0<x<4. Many factors affect the interval of convergance, and it is rarely symmetrical(1 vote)

- How did he come to decide that the absolute value of (x/5) is less than 1? Would any number work? Should I just assume that the absolute value of the sum of the series is less than 1?(1 vote)
- In order for a sequence to converge the r value must be less than 1 so in this case x/5 would need to be less than 1(2 votes)

## Video transcript

- [Instructor] So we have
an infinite series here, and the goal of this video is to try to figure out the interval of convergence for this series. And that's another way of
saying for what x values, what range of x values, is this series going to converge? And like always pause this video and see if you can figure it out. When you look at this series, it doesn't fit cleanly into something like a geometric series
or an alternating series. When I see something like this, I think about the ratio test, because it tends to be pretty general. To apply the ratio test, we want to think about the limit, the limit as n approaches infinity of the n plus oneth term, divided by the nth term and the absolute value of that. If this thing is less than one, so when this thing is less than one, then we are going to converge. And the x values that make
this thing greater than one, we are going to diverge. And the x values that
make this equal to one, well then we're going to be inconclusive and so we're going to have
to use other techniques to think about whether we're
going to converge or diverge. So let's just think about
this, let's just evaluate this, so let's do that. Limit as n approaches infinity of the absolute value of a sub-n plus one well that's going to
be x to the n plus one. Let me color code this, just
so we know what we're doing. So this thing right over
here is going to be x to the n plus one over n plus one times five to the n plus one. We're going to divide that by the nth term. We're going to divide
that by the nth term. And that's just going to be x to the n over n times five to the n. We're going to take the absolute
value of this whole thing. Now let me simplify this. I'll simplify it down here. So this is the same thing as x to the n plus one over n plus one times five to the n plus one, times the reciprocal of this. So it's going to be n times five to the n, over x to the n. And we could simplify this. This is going to be equal to, let's see, you divide
numerator and denominator by x to the n, you're left with just an x. Then divide numerator and denominator by five to the n. That is going to be,
that's a one, this is a one and then this is just going to be, five to the n plus one
divided by five to the n, that's just going to be a five. And so what do we have? We have x times n. x times n, over, distribute the five, five n, five n plus one. Oh, let me be careful there. Let me distribute the five. Five n plus five, five times n, five times one. Five n plus one. Five n plus five (chuckles). All right, so let me just rewrite that. This is going to be equal to the limit as n approaches infinity of the absolute value of this thing. And actually to help us with this limit, let me rewrite it a little bit. Let me divide the numerator
and the denominator both by n. I'm not changing the value,
I'm doing the same thing to the numerator and the denominator. I'm dividing it by the same value. So if I divide the numerator
and the denominator by n, this is going to be the same thing as x over five plus five to the n. So when you divided the numerator
and the denominator by n, it becomes very clear what
happens as n approaches infinity, As n approaches infinity,
x doesn't change, five doesn't change, but five over n goes to zero. And so this limit is going
to be equal to x over five. So that's a pretty neat, clean thing. Now we can use this to think about, and actually let me write this. This is going to be the
absolute value of x over five. Now we can think about
under which conditions is the absolute value of x over five going to be less than one and we're definitely gonna converge? Under what conditions are we going to be greater than one and definitely diverge? And then under what
conditions is it inconclusive? So let's just see when
we know we can converge. So the absolute value of x over five is less than one. This is our convergence situation. Well, that's the same thing as saying that x negative one is less than x over five, which is less than one. And you multiply all the sides by five. This is the same thing as negative five is less than x, which is less than five. So if we know that this is true, this is definitely going to be part of the interval of convergence. We know that if x meets these constraints, then our series is going to converge. But we're not done yet. We have to think about
the inconclusive case. So let's think about the scenario where the absolute value of x over five, absolute value of x over
five is equal to one. Or another way of thinking about this, that means x over five is equal to one or x over five is equal to negative one. And this means that x is equal to five or x is equal to negative five. So these are the two inconclusive cases using the ratio test. So let's test them out individually by looking back at the series and just substituting x equals five or x equals negative five. So in the first scenario, let me find a new color
here, let me use red. So the first scenario of x equals five, let's go to our series. Then the series is going to be the sum from n equals one to infinity of five to the n over n times five to the n. Well, this is just going
to be equal to the sum n equals one to infinity of one over n. This is a harmonic series. This is the p-series
where p is equal to one. And we know our p-series
of p is equal to one. That's going to diverge. And we know the harmonic series
we've done in other videos, this definitely diverges. So this diverges. You could do that by
p-series convergence test. If the p for a p-series is one, well you're gonna diverge. So now let's think about, so five is definitely not part of our interval of convergence. Now let's think about
x equals negative five. When x equals negative five, let me get a another color going here. When x is equal to negative five, then this thing is going to be equal to the sum from n equals one to infinity of negative five to the n. Actually let me just write that as, I'll write it out negative five to the n over n times five to the n. This is the same thing as the sum from n equals one to infinity. We could write this as
negative one to the n times five to the, times five to the n, over n times five to the n. And now this thing, this is an alternating harmonic series. And so you could actually use the, you might already know
that that converges, or you could use the
alternating series test. The alternating series test, it might be a little bit
clearer if I write it like this. That this is an alternating series. So in an alternating series test, if we see that this thing
is monotonically decreasing and the limit as n
approaches infinity is zero, this thing converges. The alternating harmonic
series actually converges. So this converges. So given that this converges, you could view this as this boundary here. We would include that in
our interval of convergence. So x doesn't just have to be strictly greater than negative five, it could be greater than
or equal to negative five, but it has to be less than five. This is our true interval of convergence.