Question

Help with finding the interval of convergence.....

Answer 1

\[\sum_{n=0}^{\infty}\frac{ (-1)^{n}(x+1)^{n} }{ 2^{n}}\] let me explain my problem.

Answer 2

\[\lim_{n\to inf}\left|\frac{x+1}{2}\right|=\left|\frac{x+1}{2}\right|\]
\[|\frac{x+1}{2}|<1\]
\[|x+1|<2\]
\[|x|<1\]

Answer 3

the textbook examples (this is one of them) say to use the ratio test to find the radius of convergence, then plug the endpoints into the original series to find endpoint convergence. When using the ratio test for this series, You arive at the limit being\[\left| \frac{ x+1 }{ 2 } \right|\] then what has always been done before in their examples is to set this generic limit less than one, because per the ratio test anything less than one converges. This time they set the limit less than one and solved till it said \[\left| x+1 \right|<2 \] and stopped there. so now they say the radius of convergence is R=2 and it is centered at x=-1. They did not do that in previous examples.... please help me understand how to do these, I Am having to learn these on my own

Answer 4

@amistre64 that is exactly how they have done all the other examples yet this time they did not

Answer 5

hmmm, im thinking its most likely due to a horizontal shift in a function doesnt change the properties of the function, only its position

Answer 6

|x+1|<2

-2 < (x+1) < 2

-3 < x < 1

Answer 7

Then I watched a dvd series that goes with the textbook and the guy on there said to find the radius of convergence you actually use something that looks like the ratio test, but that is not. In his example you use this formula\[R=\lim_{n \rightarrow \infty}\left| \frac{ a _{n} }{ a _{n+1} } \right|\]