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## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 16: Representing functions as power series

# Interval of convergence for derivative and integral

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.D (LO)
,
LIM‑8.D.6 (EK)
,
LIM‑8.G (LO)
,
LIM‑8.G.1 (EK)
Integrating or differentiating a power series term-by-term can only work within the interval of convergence. The interval of convergence of the integral/derivative will be the same, except maybe for the endpoints. See an example here.

## Want to join the conversation?

• I'm a bit confused at .

I understand how (1)^n-1 will diverge as it simply 1+1+1+1+1.

But for (-1)^n-1: That comes out to 1+ (-1) + 1 + (-1) + 1 + (-1) + (1) + (-1) + (1) + ...
If we re-express that as: 1+ [(-1) + 1] + [(-1) + 1] + [(-1) + (1)] + [(-1) + (1)] + ...
Shouldn't the remaining values cancel out leaving the series to converge to 1?
Am I approaching this incorrectly? • Why did you choose to cancel out that way?
Why not (1 - 1) + (1 - 1) + (1 - 1) + ... = 0 ?

Or why not this?
S = 1 - 1 + 1 - 1 + ...
+ S : 1 - 1 + - 1 + ...
=> 2S = 1
So S = 1/2

Or countless other re-arrangements?

You're not "approaching this incorrectly", you are making an incorrect assumption. Namely that infinite sums behave the same way as finite ones. Specifically with regard to the associativity of addition; we can no longer put the parentheses where we like and necessarily get the same result.

If we take the partial sums of (-1)^(n-1) we get
1, 0, 1, 0, 1...
We say a series is convergent on a value if having got within a certain "neighbourhood" of that value, it never goes outside that neighbourhood again. And that for any given neighbourhood there are only a finite number of terms in it.
So your series doesn't converge on 1, because its first term is 1, but the next term (0) is further away.
Nor does it converge on zero, for essentially the same reason.
We can't even say it converges on 1/2, since even after an infinite number of terms it would still be in the neighbourhood [0,1].
• How do we actually check the integral one for when x=1 and x=-1? I couldn't find a familiar pattern like in the derivative one. I found that when x=1 it's Σ1/(n+1)n how do we chack for convergence in here? • For the interval of convergence for the integral at the endpoints, are we really looking at a p series? Or do you actually use the comparison test with 1/p^2? I just want to make sure I'm not making the problem too difficult. :) • At Sal says series centered at 0. Can any series have a center? I thought only Maclaurin and therefore Taylor had that. Also can you explain to me what does having a center mean?
(1 vote) • When doing the ratio test for the integral why hasn't he included the n power beside the (n+1) in the integral function? This would cancel out with the n denominator of the original function and give the limit to equal zero as it would be x/n+1.
(1 vote) • When you take the sum of infinite amount of definite integrals, each will have it's own C. You could say that some constant + another constant is also a constant, but how do you know that this series of constants won't make the entire series diverge? In other words, why can you pull the C out of the sum?
(1 vote) • Can you figure out what function it represents (in the IOC)?
(1 vote) • This video starts very abruptly. Why? 