Using the Ratio Test to see if a series converges or diverges?

[lmstaples]

Homework Statement

Use the Ratio Test for series to determine whether each of the following series converge or diverge. Make Reasoning Clear.

(a) $\sum^{∞}_{n=1}\frac{3^{n}}{n^{n}}$

(b) $\sum^{∞}_{n=1}\frac{n!}{n^{\frac{n}{2}}}$

Homework Equations

$if\:lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}}) < 1 \Rightarrow\sum^{∞}_{n=1}\:-\:converges$

$if\:lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}})\in[0,1)\Rightarrow \sum^{∞}_{n=1}\:-\:diverges$

$0\leq \sum^{∞}_{n=1}(a_n)\leq \sum^{∞}_{n=1}(b_n)$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\vdots$
$if\:\sum^{∞}_{n=1}(a_n)\:-\:diverges\:\Rightarrow\:\sum^{∞}_{n=1}(b_n)\:-\:diverges$

$if\:\sum^{∞}_{n=1}(b_n)\:-\:converges\:\Rightarrow\:\sum^{∞}_{n=1}(a_n)\:-\:converges$

The Attempt at a Solution

(a) I let $a_{n}$ = sequence of partial sums then plugged everything into ratio test formula.

I ended up with:
$lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}})\:=\:3lim_{n\rightarrow∞}(\frac{n^{n}}{(n+1)(n+1)^{n}})$

I know that the limit equals 0 hence the series converges but not quite sure how to show that the limit cancels down to show 0 is "obvious"

Any help would be great, thanks.

(b) I let $a_{n}$ = sequence of partial sums then plugged everything into ratio test formula.

I ended up with:
$lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}})\:=\:lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:+\:lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})$

I know that:

$lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:=\:∞$

And that:

$lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})\:=\:0$

Hence the overall limit = ∞ and so the series diverges; but, once again I'm not quite sure how to show that how the limits cancels down to show ∞ and 0 is "obvious"

Once again, any help would be great, thanks.

 

[LCKurtz]

Homework Statement

Use the Ratio Test for series to determine whether each of the following series converge or diverge. Make Reasoning Clear.

(a) $\sum^{∞}_{n=1}\frac{3^{n}}{n^{n}}$

(b) $\sum^{∞}_{n=1}\frac{n!}{n^{\frac{n}{2}}}$

Homework Equations

$if\:lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}}) < 1 \Rightarrow\sum^{∞}_{n=1}\:-\:converges$

The Attempt at a Solution

(a) I let $a_{n}$ = sequence of partial sums then plugged everything into ratio test formula.

I ended up with:
$\lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}})=3\lim_{n\rightarrow∞}(\frac{n^{n}}{(n+1)(n+1)^{n}})$

I know that the limit equals 0 hence the series converges but not quite sure how to show that the limit cancels down to show 0 is "obvious"

The ratio test doesn't use the sequence of partial sums. It just compares $a_{n+1}$ with $a_n$. Your ratio of$$
\frac{3n^n}{(n+1)(n+1)^{n}}$$ is the right calculation. What happens if you overestimate the $n^n$ in the numerator with $(n+1)^{n}$?

(b) I let $a_{n}$ = sequence of partial sums then plugged everything into ratio test formula.

I ended up with:
$lim_{n\rightarrow∞}(\frac{a_{n+1}}{a_{n}})\:=\:lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:+\:lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})$

I know that:

$lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:=\:∞$

And that:

$lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})\:=\:0$

Hence the overall limit = ∞ and so the series diverges; but, once again I'm not quite sure how to show that how the limits cancels down to show ∞ and 0 is "obvious"

Once again, any help would be great, thanks.

For (b) I get a ratio of$$
(n+1)^{\frac 1 2}\left(\frac n {n+1}\right)^{\frac n 2}$$The right half of that has something to do with the number $e$. Is that enough of a hint?