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Determine if series 3/n*sqrt(n) converges or diverges. 
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#1
Oct809, 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 pseries 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. 


#2
Oct809, 12:40 AM

P: 867

Your b_{n} 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 n^{1 + 1/n} as n→∞ 


#3
Oct809, 01:04 AM

Mentor
P: 21,312

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 [tex]\lim_{n \rightarrow \infty}\frac{a_n}{b_n}~=~0[/tex] (with a_{n} and b_{n} 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/(n^{1/n}) = 3, and not 0 as you wrote. This shows that your series diverges, which is in agreement with your instincts. 


#4
Oct809, 08:49 PM

P: 61

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