I can’t recall if this was ever something I learned in a math course, or I’m just imagining it, but is there a generalized foolproof way to determine the convergence of a series? I know of plenty of tests, but what happens when you can’t get them to work?
(I may as well mention the series I can’t figure out - the nth term is 1/ (ln n)^(ln n) n>2. Probably some easy comparison but I don’t see it ).
Hey now, is this a homework assignment?
I’m afraid there is no magic test for telling you whether any general series, any one at all, is convergent. There are just the several tests that you’re familiar with. If you can’t match your series to one or more of the tests, then all you can do is shrug in frustration and try to appreciate life’s great mysteries, of which your series is yet another.
Maybe this approach will work?
[ul]
[li]Prove Sum 1/(ln n) is convergent.[/li]
[li]Prove that each term 1/(ln n)^(ln n) is less than 1/(ln n). (Could be tough.)[/li]
[li]The “Comparison Test” says that if you have a convergent series A of non-negative terms, and another series B of non-negative terms, and every B term is less than or equal to the corresponding A term, then series B is also convergent.[/li]
[li] Q.E.D.[/li][/ul]
Up to you though to see if this actually works.
1/ln(n) is divergent, so that’s no good.
ln(n)[sup]ln(n)[/sup] is asymptotically bounded above by n, so you have a divergent series.