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[answered]

I want to know why this particular approach is wrong so I can learn from my mistakes.

[tex]a_n = \frac{ln(n^3)}{2n}[/tex]

For the sake of being time efficient, I will skip writing things like the limit as n approaches infinity etc.

[tex]a_n = \frac{ln(n^3)}{2n}[/tex]

I let n^3 = u

[tex]a_n = \frac{ln(u)}{2n}[/tex]

Using L'hopital's rule,

[tex]\frac{3n^2}{2n^3}[/tex]

Applying it once more,

[tex]\frac{6n}{6n^2}[/tex]

Which simplifys to

[tex]\frac{6}{12n}[/tex]

This shows that the function diverges. Which is the wrong answer.

I want to know why this particular approach is wrong so I can learn from my mistakes.

## Homework Statement

[tex]a_n = \frac{ln(n^3)}{2n}[/tex]

## The Attempt at a Solution

For the sake of being time efficient, I will skip writing things like the limit as n approaches infinity etc.

[tex]a_n = \frac{ln(n^3)}{2n}[/tex]

I let n^3 = u

[tex]a_n = \frac{ln(u)}{2n}[/tex]

Using L'hopital's rule,

[tex]\frac{3n^2}{2n^3}[/tex]

Applying it once more,

[tex]\frac{6n}{6n^2}[/tex]

Which simplifys to

[tex]\frac{6}{12n}[/tex]

This shows that the function diverges. Which is the wrong answer.

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