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

**lmstaples** · November 22nd 2012, 03:30 PM

1. The problem statement, all variables and given/known data

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

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

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

2. Relevant equations

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

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

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

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

3. 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:
$lim_{n\rightarrow infinity}(\frac{a_{n+1}}{a_{n}})\:=\:3lim_{n|:tend s\:to\:infinity}(\frac{n^{n}}{(n+1)(n+1)^{n}})$

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:
$lim_{n\rightarrow}(\frac{a_{n+1}}{a_{n}})\:=\:lim_ {n\rightarrow infinity}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\s qrt{n+1}})\:+\:lim_{n\rightarrow infinity}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n} {2}}\sqrt{n+1}})$

I know that:

$lim_{n\rightarrow infinity}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\s qrt{n+1}})\:=\:infinity$

And that:

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

Hence the overall limit = infinity 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.

**Prove It** · November 22nd 2012, 05:34 PM

For a) notice that $\displaystyle \begin{align*} a_n = \frac{3^n}{n^n} \end{align*}$ and $\displaystyle \begin{align*} a_{n+1} = \frac{3^{n+1}}{(n + 1)^{n+1}} \end{align*}$ . Then

$\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{\frac{3^{n+1}}{(n+1)^{n+1}}}{\frac{3^n}{n^n} } \right| \\ &= \lim_{n \to \infty} \frac{3^{n+1}\, n^n}{3^n\left (n + 1 \right)^{n+1}} \\ &= \lim_{n \to \infty} \left[ \frac{3}{(n+1)} \left( \frac{n}{n+1} \right)^n \right] \\ &= \lim_{n \to \infty} \frac{3}{n+ 1} \cdot \lim_{n \to \infty} \left[ \left( \frac{n}{n+1} \right)^n \right] \\ &= 0 \lim_{n \to \infty} e^{\ln{\left[ \left( \frac{n}{n+1} \right)^n \right]}} \\ &= 0 \, e^{ \lim_{n \to \infty} \frac{\ln{\left( 1 - \frac{1}{n+1} \right)}}{\frac{1}{n}} } \\ &= 0\, e^{\lim_{n \to \infty} \frac{\frac{1}{n} - \frac{1}{n+1}}{-\frac{1}{n^2}}} \textrm{ by L'Hospital's Rule} \\ &= 0 \, e^{\lim_{n \to \infty} \frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}}} \\ &= 0\, e^{\lim_{n \to \infty} -\frac{n^2}{n^2 + n}} \\ &= 0\, e^{\lim_{n \to \infty} -\frac{1}{1 + \frac{1}{n}}} \\ &= 0 \, e^{-1} \\ &= 0 \end{align*}$

**lmstaples** · November 22nd 2012, 05:45 PM

Thanks, that's very good and I understand all of it; however, I haven't been formally taught L'Hospitals Rule yet and so am not allowed to use it for this particular problem.

Any other ideas?

**Prove It** · November 22nd 2012, 05:52 PM

Say that it's self-taught then.

**lmstaples** · November 22nd 2012, 05:55 PM

Could I equally get away with saying that since the first part of the limit = 0 then it doesn't matter what the second part = since it will still be zero?

**Prove It** · November 22nd 2012, 05:59 PM

Originally Posted by lmstaples

Could I equally get away with saying that since the first part of the limit = 0 then it doesn't matter what the second part = since it will still be zero?

No, because $\displaystyle \begin{align*} 0 \times \infty \end{align*}$ is an entirely different kettle of fish.

I expect you would have to make use of $\displaystyle \begin{align*} \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e \end{align*}$ somehow...

**lmstaples** · November 22nd 2012, 06:02 PM

Now that standard limit I have been taught and can use. I was trying to implement it before buy didn't get anywhere - I should have kept with it longer.

**Soroban** · November 22nd 2012, 07:00 PM

Hello, Imstaples!

Another approach . . .

$\displaystyle \sum^{\infty}_{n=1}\,\frac{3^n}{n^n}$

$\text{Ratio} \;=\;\frac{3^{n+1}}{(n+1)^{n+1}}\cdot\frac{n^n}{3^ n} \;=\; \frac{3^{n+1}}{3^n}\cdot\frac{n^n}{(n+1)^{n+1}} \;=\;3\cdot\frac{n^n}{(n+1)(n+1)^n}$

. . . . . $=\;\frac{3}{n+1}\cdot\frac{n^n}{(n+1)^n} \;=\;\frac{3}{n+1}\left(\frac{n}{n+1}\right)^n \;=\;\frac{3}{n+1}\left(\frac{1}{1+\frac{1}{n}} \right)^n$

. . . . . $=\;\frac{3}{n+1}\cdot\frac{1}{(1+\frac{1}{n})^n}$

$\displaystyle \lim_{n\to\infty} \frac{3}{n+1}\cdot\frac{1}{(1+\frac{1}{n})^n} \;=\; \underbrace{\lim_{n\to\infty}\frac{3}{n+1}}_{\text {This is 0}} \cdot\frac{1}{\underbrace{\lim_{n\to\infty}\left(1 +\tfrac{1}{n} \right)^n}_{\text{This is }e}} \;=\;0\cdot\frac{1}{e} \;=\;0$

**MarkFL** · November 22nd 2012, 07:53 PM

To the OP, I like very much the way you presented the problem.

Let me offer a small tip for your use of $\LaTeX$ :

i) Use the code \infty to get $\infty$

ii) Use the code \lim_{n\to\infty} to get $\lim_{n\to\infty}$

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