# Math Help - Interval of Convergence

1. ## Interval of Convergence

I'm trying to find the interval of convergence:

$f(x) = \sum^ \infty_ {n=0} \frac{(-1)^{n+1}(x-1)^{n+1}}{n+1} = \mid x-1 \mid$

I still don't know how to find its interval of convergence. Could you explain the process to me? I really want to understand this concept before my final test on Monday. Thank you.

2. Originally Posted by saiyanmx89
I'm trying to find the interval of convergence:

$f(x) = \sum^ \infty_ {n=0} \frac{(-1)^{n+1}(x-1)^{n+1}}{n+1} = \mid x-1 \mid$

I still don't know how to find its interval of convergence. Could you explain the process to me? I really want to understand this concept before my final test on Monday. Thank you.
I have a counter question for you. It seems to me that you are using the ratio test. What is the statement of the ratio test and what does it tells us to do to $|x-1|$ I think that this is what you are having problems with. You seems to know what to do algorithmically, but I'm not sure you understand what the conclusion of the ratio test is. In your own words please state the conclusion of the ratio test.

3. Assume that $\lim_ {n\rightarrow \infty} \mid \frac{a_n+1}{a_n}\mid = L$
If L <1, then series is abs. conv.
If L > 1, then series is div.
If L = 1, then test fails.

4. Originally Posted by saiyanmx89
Assume that $\lim_ {n\rightarrow \infty} \mid \frac{a_n+1}{a_n}\mid = L$
If L <1, then series is abs. conv.
If L > 1, then series is div.
If L = 1, then test fails.
Okay that is correct for the statement, but what is it asking us to do.

It is telling us that we need to know when

$|x-1|< 1 \iff -1 < x-1 < 1 \iff 0 < x < 2$

So we know that the series converges for the above values of x. Note that the test fails when L=1 so we have to check the endpoints on an individual basis. That means we need to plug x=0 and x=2 back into the series and analyze what happens.

I will do one of them and leave the other for you.

when x=0 we get

$\sum^ \infty_ {n=0} \frac{(-1)^{n+1}(0-1)^{n+1}}{n+1}=\sum^ \infty_ {n=0} \frac{(-1)^{n+1}(-1)^{n+1}}{n+1}=\sum^ \infty_ {n=0} \frac{(-1)^{2n+2}}{n+1}=\sum^ \infty_ {n=0} \frac{1}{n+1}$

This series diverges by the integral test, or you can note that it is the tail end of the harmonic series.

You will need to check the other one.

I hope this helps. Good luck