## AmTran_Bus one year ago Interval of convergence Check work

1. AmTran_Bus

|dw:1438630716254:dw|

2. AmTran_Bus

since abs(x-6) <1?

3. anonymous

$$\frac{\sqrt[4]{n+1}\,(x-6)^{n+1}}{\sqrt[4]n\ (x-6)^n}=\sqrt[4]{1+\frac1n}\cdot (x-6)$$so the ratio test shows $$\lim_{n\to\infty}\sqrt[4]{1+\frac1n}\cdot \left|x-6\right|=|x-6|<1$$ so we know the radius of convergence is $$R=1$$. now we test the boundary points: for $$x=5$$ we have $$\sum_{n=0}^\infty \sqrt[4]{n}(-1)^n$$ since $$(-1)^n\sqrt[4]{n}\not\to 0$$ it diverges. similarly for $$x=7$$ $$\sum_{n=0}^\infty\sqrt[4]{n}$$ again, $$\sqrt[4]\not\to 0$$ so it diverges

4. anonymous

again, $$\sqrt[4]{n}\not\to 0$$ so it diverges * i meant to say

5. AmTran_Bus
6. AmTran_Bus

@oldrin.bataku

7. AmTran_Bus

Because I really thought (5,7) was right after solving the inequality.

8. freckles

He was agreeing with you it diverges at the endpoints 5 and 7.

9. freckles

which means it is (5,7) not [5,7] or [5,7) or (5,7]

10. AmTran_Bus

thanks @freckles