Divergent Series - How to test it

I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.

 

True, but 1/n is on the boundary between divergence and convergence for power series. You need an actual comparison to make. Being asymptotically 'like' 1/n is not good enough. Choose a series with a smaller denominator and a larger denominator and show it diverges. I'm a little worried that what Saterial has said so far makes little sense.

 

Yes, that would be incorrect.

No. It isn't a p series

Look up the "general comparison test" or it may be called the "limit comparison test" in your text.

 

That would be a correct intuitive argument, but you need the limit comparison test to prove it.

 

A direct comparison test will do it. And it's simpler.

 

Probably so and point taken. I haven't actually worked on it. But I do think it is good for students to recognize situations that are a natural setting for the limit comparison test.

 

I'll repeat what I said before. Make the denominator smaller and the numerator larger in a way that the resulting series still diverges and you've got it. Limit comparison test may not be in Saterial's repetoire. And it's not needed. It's really simple once you've seen it.

 

Making the denominator smaller and the numerator larger (and hence every term larger) would not help you prove divergence.

Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?

 

Yes, the divergence follows rather trivially when one does that because the series collapses into the harmonic series.

 

(1+n+n²)/(√1+n²+n⁶)) > (1)/(√1+n²+n⁶)) > (1)/(√n²+n²+n⁶)) = (1)/(√2n²+n⁶)) = (1)/(√n²(2+n⁴))) = (1)/(n√(2+n⁴))) > 1/n

 

Oh yeah. That's what I meant. Denominator is the thing on the BOTTOM. Duh. Thanks!

 

Sorry. (1)/(n√(2+n⁴))) is less than 1/n. Not greater.

 

How about just noting that the original series is greater than n^2/sqrt(n^6+n^6+n^6)?

 

Yes, that would work. Much simpler, too.

I just showed the first thing I thought of.

 

Or as another alternative noting that it's greater than [itex]\frac{n^2}{n^3 + 1}[/itex], which in turn is greater than [itex]\frac{1}{n+1}[/itex]. We can then compare it directly to the harmonic series starting at n=2.