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


by NastyAccident
Tags: 3 or nsqrtn, converges, determine, diverges, series
NastyAccident
NastyAccident is offline
#1
Oct8-09, 12:27 AM
P: 61
1. The problem statement, all variables and given/known data

[tex]\sum^{\infty}_{n=1}\frac{3}{n^{1+\frac{1}{n}}}[/tex]

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


3. The attempt at a solution
I sort of just would like to check to make sure I'm getting a proper [tex]b_{n}[/tex]

Manipulating the Series:
[tex]\sum^{\infty}_{n=1}\frac{3}{n^{1+\frac{1}{n}}}[/tex]

[tex]\sum^{\infty}_{n=1}\frac{3}{n^{\frac{n+1}{n}}}[/tex]

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

Test of Convergence:
limit n->infinity [tex]\frac{3}{n^{\frac{n+1}{n}}}[/tex]

limit n->infinity [tex]\frac{3}{n^1}}[/tex]

limit ->infinity [tex]0[/tex]

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.
[tex]a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}[/tex]
[tex]a_{n} \geq b_{n}[/tex]

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

By the Comparison Test, [tex]\sum^{\infty}_{n=1}\frac{3}{n\sqrt[n]{n}}[/tex] 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*
[tex]a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}[/tex]

limit n->infinity [tex]\frac{\frac{3}{n^{\frac{n+1}{n}}}}{\frac{1}{n}}[/tex]

limit n->infinity [tex]\frac{3n}{n^{\frac{n+1}{n}}}}[/tex]

limit n->infinity [tex]\frac{3}{n^{\frac{1}{n}}}}[/tex]

limit n->infinity [tex]0[/tex]


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

My questions:
Did I pick the right [tex]b_{n}[/tex]? If not, what did I do wrong in picking [tex]b_{n}[/tex]?
Any hints for picking the proper [tex]b_{n}[/tex]?
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!])

Sincerely,

NastyAccident.
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Bohrok
Bohrok is offline
#2
Oct8-09, 12:40 AM
P: 867
Your bn seems great for your tests.

[tex]\lim_{n\rightarrow\infty}\frac{3}{n^{\frac{n+1}{n}}}[/tex] does have a limit; look at the denominator n1 + 1/n as n→∞
Mark44
Mark44 is offline
#3
Oct8-09, 01:04 AM
Mentor
P: 21,063
For your comparison test, with an being the terms in your series, and bn being the terms in the harmonic series, you said that an >= bn. 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
[tex]\lim_{n \rightarrow \infty}\frac{a_n}{b_n}~=~0[/tex]
(with an and bn still as defined above), which is not true.

In your work you show [tex]\frac{\frac{3}{n^{1 + 1/n}}}{\frac{1}{n}}~=~\frac{3}{n^{1/n}}[/tex]
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:
[tex]Let~y~=~n^{1/n}[/tex]
[tex]Then~ln~y~= ln (n^{1/n})~=~1/n*ln~n~=~\frac{ln~n}{n}[/tex]
Taking the limit of both sides, we have
[tex]\lim_{n \rightarrow \infty}ln~y~=~\lim_{n \rightarrow \infty}\frac{ln~n}{n}[/tex]
[tex]=~\lim_{n \rightarrow \infty}\frac{1/n}{1}~=~0[/tex]
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/(n1/n) = 3, and not 0 as you wrote.

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

NastyAccident
NastyAccident is offline
#4
Oct8-09, 08:49 PM
P: 61

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


Quote Quote by Mark44 View Post
For your comparison test, with an being the terms in your series, and bn being the terms in the harmonic series, you said that an >= bn. 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
[tex]\lim_{n \rightarrow \infty}\frac{a_n}{b_n}~=~0[/tex]
(with an and bn still as defined above), which is not true.

In your work you show [tex]\frac{\frac{3}{n^{1 + 1/n}}}{\frac{1}{n}}~=~\frac{3}{n^{1/n}}[/tex]
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:
[tex]Let~y~=~n^{1/n}[/tex]
[tex]Then~ln~y~= ln (n^{1/n})~=~1/n*ln~n~=~\frac{ln~n}{n}[/tex]
Taking the limit of both sides, we have
[tex]\lim_{n \rightarrow \infty}ln~y~=~\lim_{n \rightarrow \infty}\frac{ln~n}{n}[/tex]
[tex]=~\lim_{n \rightarrow \infty}\frac{1/n}{1}~=~0[/tex]
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/(n1/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):
[tex]a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}[/tex]

[tex]a_{n} \geq b_{n}[/tex]

[tex]\frac{3}{n^{1+\frac{1}{n}}}\geq\frac{1}{n}[/tex]

[tex]\frac{3}{1^{1+\frac{1}{1}}}\geq\frac{1}{1}[/tex]

[tex]\frac{3}{2^{1+\frac{1}{2}}}\geq\frac{1}{2}[/tex]

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.):

Indeterminate form
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!

Sincerely,

NastyAccident


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