Question

See post

Answer 1

\[\sum_{n=1}^{\infty}\frac{(\sqrt{n+1})(x-3)^n}{5^n}\]

Determine the radius of convergence and the interval of convergence.

Just want to check if I did it correctly on my final.

I got

Radius=5

Interval\[-2<x <8\]

Where\[\sum_{n=1}^{\infty}\frac{(\sqrt{n+1})(-2-3)^n}{5^n}\]Was divergent by the absolute convergence theorem

And\[\sum_{n=1}^{\infty}\frac{(\sqrt{n+1})(8-3)^n}{5^n}\]Was divergent by the test for divergence.

Answer 2

i've not worked out the entire problem...but to find interval of convergence you need to do the ratio test to get the endpoints...then you check the endpoints for your intervals as to whether or not to include them. For the radius of convergence the formula is b-a / 2...if your interval endpoints are -2 and 8 it would be 8 - (-2) / 2 which is 5 so you have that correct.