Good morning All,

I am working on ratio and root tests for infinite series. I would like to post what I have done, and see if anyone agrees with it. Please see below:

$\displaystyle

\displaystyle\sum_{n=2}^{\infty}(-1)^n\frac{123^n}{n!}

$

By the Ratio Test...

$\displaystyle

\displaystyle\lim_{n\rightarrow\infty}\mid\frac{(-1)^{(n+1)}123^{(n+1)}}{(n+1)!}\frac{n!}{(-1)^{n}123^{n}}\mid

$

And...

$\displaystyle

\displaystyle\lim_{n\rightarrow\infty}\frac{123n!} {(n+1)!} = \lim_{n\rightarrow\infty}\frac{123n!}{(n+1)n!} = \lim_{n\rightarrow\infty}\frac{123}{n+1} = 0

$

Therefore...

The orignal series $\displaystyle \displaystyle\sum_{n=2}^{\infty}(-)^n\frac{123^n}{n!}$ isAbsolutely Convergent.

Are we good here?

My second problem is:

$\displaystyle \displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$

I'm not really sure where to start here. Would I apply the root test to this problem?