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

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Homework Statement



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

(a) [itex]\sum^{∞}_{n=1}\frac{3^{n}}{n^{n}}[/itex]

(b) [itex]\sum^{∞}_{n=1}\frac{n!}{n^{\frac{n}{2}}}[/itex]


Homework Equations



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

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

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

[itex]if\:\sum^{∞}_{n=1}(b_n)\:-\:converges\:\Rightarrow\:\sum^{∞}_{n=1}(a_n)\:-\:converges[/itex]

The Attempt at a Solution



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

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

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 [itex]a_{n}[/itex] = sequence of partial sums then plugged everything into ratio test formula.

I ended up with:
[itex]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}})[/itex]

I know that:

[itex]lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:=\:∞[/itex]

And that:

[itex]lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})\:=\:0[/itex]

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.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



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

(a) [itex]\sum^{∞}_{n=1}\frac{3^{n}}{n^{n}}[/itex]

(b) [itex]\sum^{∞}_{n=1}\frac{n!}{n^{\frac{n}{2}}}[/itex]


Homework Equations



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

The Attempt at a Solution



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

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

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 [itex]a_{n}[/itex] = sequence of partial sums then plugged everything into ratio test formula.

I ended up with:
[itex]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}})[/itex]

I know that:

[itex]lim_{n\rightarrow∞}(\frac{n(\frac{n}{n+1})^{\frac{n}{2}}}{\sqrt{n+1}})\:=\:∞[/itex]

And that:

[itex]lim_{n\rightarrow∞}(\frac{(n)^{\frac{n}{2}}}{(n+1)^{\frac{n}{2}}\sqrt{n+1}})\:=\:0[/itex]

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?
 

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