1/(lnk)^3

: Math 2730 Sequences and Series: 1/(lnk)^3
   By Martin Daniels on Wednesday, February 27, 2008 - 12:21 pm: Edit Post

I'm working on the series

sigma(k=1, inf) 1/(lnk)^3

The first thing I wanted to try was to compare it with a similar series, but I couldn't find anything that suited it well.

Next I thought about the limit comparison test, but again I didn't have anything suitable.

Finally, and perhaps the best option, I wanted to try the integral test for f(x) = 1/(lnx)^3 but I had no idea how to integrate that.

Any tips on how one should begin a problem of this nature?


   By Sasho on Wednesday, February 27, 2008 - 02:46 pm: Edit Post

One way to do it would be to compare with the series sigma (1/k).
The point is that (lnx)3 is less than x from some point on. That implies that 1/(lnx)3>1/x from some point on, and so 1/(lnk)3>1/k from some point on. Since sigma (1/k) diverges to infinity, so does sigma 1/(lnk)3.

Now, showing (lnx)3<x for large enough x is the same as showing lnx<x1/3 for large enough x, and that could be done using calculus 1 by analyzing that function using derivatives.


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