Interval of Convergence and radicals

[justtip]

Homework Statement

Find the interval of convergence:

[tex]\sum _{n=1}^{\infty } \frac{(-1)^n (x+2)^n}{3^n\sqrt{n}}[/tex]

Homework Equations

The Attempt at a Solution

[tex]\lim_{n\to \infty } |\frac{(x+2)^{n+1}}{3^{n+1}\sqrt{n+1}}*\frac{3^n\sqrt{n}}{(x+2)^n}| = \lim_{n\to \infty } |\frac{(x+2)\sqrt{n}}{3\sqrt{n+1}}|[/tex]


This is where I'm stuck. Mathematica says the limit is (x+2)/3, but I'm not sure how to get there. Only thing I can think of is if infinity is substituted for n, the square root of infinity cancels out? I can work the rest out myself, just need to know how to get to (x+2)/3.

Thanks.

 

[tiny-tim]

welcome to pf!

hi justtip! welcome to pf!

Mathematica says the limit is (x+2)/3, but I'm not sure how to get there.

you're trying to prove that lim √n/√(n+1) = 1

that's the same as lim √(n/(n+1)) = 1,

which is the same as lim n/(n+1) = 1