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Does there exist a $c$ such that the series $1/[c^2(\ln x)]$ converges? I know it's a simple question but I am working on a probability proof and if I can find such a $c$ I would be able to apply the Borel Cantelli lemmas and complete the proof. Any help would be appreciated.

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Constants won't affect convergence? – Alex Youcis Dec 12 '12 at 3:52
Your original post asked about 1/{c^2(ln x)} which lost the braces in the edit. However, this would be read as $\frac 1{c^2 \ln x}$ while you probably meant $\frac 1{c^{2 \ln x}}$. Can you confirm? – Ross Millikan Dec 12 '12 at 4:11
up vote 1 down vote accepted

If the question is about $\dfrac{1}{(c^2)(\ln x)}$ then of course not, since $\sum_{x=2}^\infty \dfrac{1}{\ln x}$ diverges.

So the question must be about $\dfrac{1}{c^{2\ln x}}$. Then sure, $c=e$ will do, indeed any $c\gt e^{1/2}$. For then we get a series $\sum \dfrac{1}{x^p}$ with $p\gt 1$, and these are known to converge.

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Yeah... The question is to show that P(|Xn|>= c(sqrt n) i.o) = 0 and to identify the smallest c. I was trying to use Chebychevs inequality and Borel-Cantelli but maybe this is not the right approach. The Xn are normally distributed. – rmh52 Dec 12 '12 at 4:05
I guess I would estimate the tail of the integral. That's fairly straighforward. For the normal I would get the estimate by integration by parts. Have seen more clever ways. – André Nicolas Dec 12 '12 at 4:11

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