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I have to prove this converges using the limit comparison test and I can't figure out what to compare to. I have tried $1/n$, $1/n^{1/2}$, $1/n^2$, $1/n^4$, $1/n^{3/2}$, and $1/n^{5/2}$.

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    Is this a series or an integral or simple a sequence?2011-11-03
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    It seems you are trying only powers of $n$, which will not work in this case. The numerator is $(\ln n)^2$ and the denominator grows like $9n^{3/2}$ for large $n$. So, how about using $(\ln n)^2/ n^{3/2}$ for comparison?2011-11-03

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