(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Determine if series 3/n*sqrt(n) converges or diverges.

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