Determine if series 3/n*sqrt(n) converges or diverges.

[NastyAccident]

Homework Statement

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

Homework Equations

Comparison Test
Limit Comparison Test
Test of Convergence (Just to show it doesn't immediately diverge)

The Attempt at a Solution

I sort of just would like to check to make sure I'm getting a proper $$b_{n}$$

Manipulating the Series:
$$\sum^{\infty}_{n=1}\frac{3}{n^{1+\frac{1}{n}}}$$

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

$$\sum^{\infty}_{n=1}\frac{3}{n\sqrt[n]{n}}$$

Test of Convergence:
limit n->infinity $$\frac{3}{n^{\frac{n+1}{n}}}$$

limit n->infinity $$\frac{3}{n^1}}$$

limit ->infinity $$0$$

The series MAY or MAY NOT be convergent.

Comparison Test
*Note* This series only contains positive terms*
From the looks of it, I'm going to GUESS that this series DIVERGES.
$$a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}$$
$$a_{n} \geq b_{n}$$

Since the series is $$\sum^{\infty}_{n=1} \frac{1}{n}$$, it is a p-series and it diverges because p $$\leq$$ 1.

By the Comparison Test, $$\sum^{\infty}_{n=1}\frac{3}{n\sqrt[n]{n}}$$ also diverges.

Limit Comparison Test

From the looks of it, I'm going to GUESS that this series DIVERGES.
*Note* This series only contains positive terms*
$$a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}$$

limit n->infinity $$\frac{\frac{3}{n^{\frac{n+1}{n}}}}{\frac{1}{n}}$$

limit n->infinity $$\frac{3n}{n^{\frac{n+1}{n}}}}$$

limit n->infinity $$\frac{3}{n^{\frac{1}{n}}}}$$

limit n->infinity $$0$$By the Limit Comparison Test, $$\sum^{\infty}_{n=1}\frac{3}{n\sqrt[n]{n}}$$ is divergent since 0 > 0.

My questions:
Did I pick the right $$b_{n}$$? If not, what did I do wrong in picking $$b_{n}$$?
Any hints for picking the proper $$b_{n}$$?
Was there a step that I missed or was unclear?

As always, any and all help is appreciated and will be greatly thanked! =) (I'm getting a 96% in Calc II thanks to the help I am receiving from this community in understanding concepts! [Nailed a 56/60 on a 20% exam!])
NastyAccident.

 

[Bohrok]

Your b_n seems great for your tests.

$$\lim_{n\rightarrow\infty}\frac{3}{n^{\frac{n+1}{n}}}$$ does have a limit; look at the denominator n^{1 + 1/n} as n→∞

 

[Mark44]

For your comparison test, with a_n being the terms in your series, and b_n being the terms in the harmonic series, you said that a_n >= b_n. That very well may be true, but you would need to establish this inequality instead of merely stating it.

For your work using the limit comparison test, you concluded that
$$\lim_{n \rightarrow \infty}\frac{a_n}{b_n}~=~0$$
(with a_n and b_n still as defined above), which is not true.

In your work you show $$\frac{\frac{3}{n^{1 + 1/n}}}{\frac{1}{n}}~=~\frac{3}{n^{1/n}}$$
which is correct, but in evaluating the denominator limit you got an incorrect value. Here's how that goes, looking just at the limit of the denominator:
$$Let~y~=~n^{1/n}$$
$$Then~ln~y~= ln (n^{1/n})~=~1/n*ln~n~=~\frac{ln~n}{n}$$
Taking the limit of both sides, we have
$$\lim_{n \rightarrow \infty}ln~y~=~\lim_{n \rightarrow \infty}\frac{ln~n}{n}$$
$$=~\lim_{n \rightarrow \infty}\frac{1/n}{1}~=~0$$
The last limit was evaluated using L'Hopital's Rule.
Since lim ln y = ln lim y = 0, this means that lim y = 1.

The upshot of all this is that lim 3/(n^1/n) = 3, and not 0 as you wrote.

This shows that your series diverges, which is in agreement with your instincts.

 

[NastyAccident]

For your comparison test, with a_n being the terms in your series, and b_n being the terms in the harmonic series, you said that a_n >= b_n. That very well may be true, but you would need to establish this inequality instead of merely stating it.

For your work using the limit comparison test, you concluded that
$$\lim_{n \rightarrow \infty}\frac{a_n}{b_n}~=~0$$
(with a_n and b_n still as defined above), which is not true.

In your work you show $$\frac{\frac{3}{n^{1 + 1/n}}}{\frac{1}{n}}~=~\frac{3}{n^{1/n}}$$
which is correct, but in evaluating the denominator limit you got an incorrect value. Here's how that goes, looking just at the limit of the denominator:
$$Let~y~=~n^{1/n}$$
$$Then~ln~y~= ln (n^{1/n})~=~1/n*ln~n~=~\frac{ln~n}{n}$$
Taking the limit of both sides, we have
$$\lim_{n \rightarrow \infty}ln~y~=~\lim_{n \rightarrow \infty}\frac{ln~n}{n}$$
$$=~\lim_{n \rightarrow \infty}\frac{1/n}{1}~=~0$$
The last limit was evaluated using L'Hopital's Rule.
Since lim ln y = ln lim y = 0, this means that lim y = 1.

The upshot of all this is that lim 3/(n^1/n) = 3, and not 0 as you wrote.

This shows that your series diverges, which is in agreement with your instincts.

Attempt to satisfy #1 (I'm assuming I do not have to do mathematical induction):
$$a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}$$

$$a_{n} \geq b_{n}$$

$$\frac{3}{n^{1+\frac{1}{n}}}\geq\frac{1}{n}$$

$$\frac{3}{1^{1+\frac{1}{1}}}\geq\frac{1}{1}$$

$$\frac{3}{2^{1+\frac{1}{2}}}\geq\frac{1}{2}$$

Attempt to satisfy #2 (Please note, I lacked knowledge on Le'Hospital's rule prior to this year since my AB teacher did not teach it.):

"[URL form
[/URL] shows that this is infinity to the 0 power case...

Transformation is:

Which matches what you did... So =) Thanks for directing me to pull out the Indeterminate forms... I'll be studying them this weekend!



NastyAccident