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Calculus, all content (2017 edition)

Course: Calculus, all content (2017 edition)>Unit 7

Lesson 12: Power series intro

Worked example: interval of convergence

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.D (LO)
,
LIM‑8.D.1 (EK)
,
LIM‑8.D.2 (EK)
,
LIM‑8.D.3 (EK)
,
LIM‑8.D.4 (EK)
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?

• 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.
• 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)
• 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?
• 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)
• 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.
• Monotonically?
(1 vote)
• Monotonically decreasing means strictly decreasing; there's no point where the n+1th term is greater than the nth term.