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.
Convergence of Series $1/[c^2(\ln x)]$
1
$\begingroup$
sequences-and-series
-
0Your 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? – 2012-12-12
1 Answers
1
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.
-
0I 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. – 2012-12-12