Testing Convergence, Calculus II

[joypav]

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[Opalg]

Unfortunately, your inequality is the wrong way round. It should say $\dfrac n{\sqrt{n^3+n}} \geq \dfrac n{n^3+n}$ (as the denominator gets larger, the fraction gets smaller!).

For this problem, notice that, as $n$ gets large, $(\ln n)^2$ will be much smaller than $n$. In the denominator, $n$ will be much smaller than $n^3$. So to a first approximation the fraction $\dfrac {n + (\ln n)^2}{\sqrt{n^3+n}}$ will look like $\dfrac n{\sqrt{n^3}} = \dfrac1{n^{1/2}}.$ Try using the limit comparison test to compare $\dfrac {n + (\ln n)^2}{\sqrt{n^3+n}}$ with $\dfrac1{n^{1/2}}.$