how do I find the convergence of $((\ln(n)^2)/(n^{1/2}(9n-10n^{1/2})$ using the limit comparison test

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? – robjohn Nov 3 '11 at 19:21
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? – Srivatsan Nov 3 '11 at 19:45

1 Answer

Assuming that you are considering the convergence of $$\sum_{n=1}^\infty\frac{\log^2(n)}{n^{1/2}(9n-10n^{1/2})}\tag{1}$$ First, you can establish that for $n\ge4$, you have $n^{1/2}\le n/2$. Therefore, $9n-10n^{1/2}\ge4n$, and thus, for $n\ge4$, each term of $(1)$ is less than the corresponding term of $$\sum_{n=1}^\infty\frac{\log^2(n)}{4n^{3/2}}\tag{2}$$ Next, show that for $n\ge e^{18}$, $\log(n)=6\log(n^{1/6})\le n^{1/6}$. This can be shown by noting that $x-6\log(x)$ is monotonic increasing for $x\ge6$, then setting $x=n^{1/6}$. Therefore, you get that for $n\ge e^{18}$, each term of $(2)$ is less than the corresponding term of $$\sum_{n=1}^\infty\frac{n^{1/3}}{4n^{3/2}}=\sum_{n=1}^\infty\frac{1}{4n^{7/6}}\tag{3}$$ and $(3)$ can be shown to converges by the integral test.

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