College Math Teaching

November 18, 2013

And I get sloppy….divergence of n!x^n

Filed under: calculus, sequences, series — Tags: — collegemathteaching @ 9:29 pm

In class I was demonstrating the various open intervals of absolute convergence and gave the usual \sum k!x^k as an example of a series that converges at x = 0 only. I mentioned that “\sum k!x^k doesn’t even pass the divergence test”, which, as it turns out, is true. But why? (yes, it is easier to just use the ratio test and be done with it)

Well, I should have noted: if x > 0 , then x > \frac{1}{m} for some integer m, then for k > m we have k!x^k > \frac{1*2*3...*m *(m+1)*(m+2)...*k}{m*m*m...*m*m*m...*m} and one can see that this is a finite number times a number which is growing without bound. Hence the sequence of terms of the series grows without bound for any positive value of x .

Advertisements

Create a free website or blog at WordPress.com.