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### Course: Integral Calculus>Unit 5

Lesson 13: Power series intro

# 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?
• 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)
• Starting from , 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.
• shouldn't we test the end points? or do we always not include them?
• 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.
• This might be a stupid question but are there any function series whose convergence intervals are not contiguous? If no, why?
• 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)
• how does 5^n/5^n+1 = 1/5
• 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!
• 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)`
• 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)
• Monotonically?
(1 vote)
• Monotonically decreasing means strictly decreasing; there's no point where the n+1th term is greater than the nth term.
• If my limit evaluates to 0 does that mean that it converges over any interval?
• 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.