Thread: Convergence problem

Ok, so I got a recent homework back for which I had to find the interval of convergence for the following series:

sigma starting at j=3 and going to infinity of:
3/(2-j) * (x+3)^j

Now, I found the interval of convergence by doing the following:

a_n+1/a_n = (n-2)/(n-1) * |x+3|

If |x+3| < 1, then the series is absolutely convergent.

-1 < x+3 < 1

Subtracting 3 from all sides, I got
-4 < x < -2, the interval of convergence.

My teacher wrote on my paper that I should check the endpoints to see if they are included. How do I go about this (my teacher went over this in class, but I seem to have lost the paper I wrote it on)?

Originally Posted by clockingly

Ok, so I got a recent homework back for which I had to find the interval of convergence for the following series:

sigma starting at j=3 and going to infinity of:
3/(2-j) * (x+3)^j

Now, I found the interval of convergence by doing the following:

a_n+1/a_n = (n-2)/(n-1) * |x+3|

If |x+3| < 1, then the series is absolutely convergent.

-1 < x+3 < 1

Subtracting 3 from all sides, I got
-4 < x < -2, the interval of convergence.

My teacher wrote on my paper that I should check the endpoints to see if they are included. How do I go about this (my teacher went over this in class, but I seem to have lost the paper I wrote it on)?

Check the the endpoints just means to substitute that into the power series and see if it converges.
Thus, for x=-2 , the right endpoint.
Substitute that for x in the power series and see if it converges.
Then do the same with x=-4.