convergent, divergent, conditionally convergent, or absolutely convergent.
$$\sum_{n=1}^{\infty} (-1)^n \frac{ (1.1)^n }{ n^4 }$$
Book says this is divergent, but does it not satisfy all the conditions for the leibniz theorem (AST) ?

Question

convergent, divergent, conditionally convergent, or absolutely convergent.
$$\sum_{n=1}^{\infty} (-1)^n \frac{ (1.1)^n }{ n^4 }$$
Book says this is divergent, but does it not satisfy all the conditions for the leibniz theorem  (AST) ?

shamil98 · Answer

Actually i just realized as n gets larger (1.1)^n approaches infinity
so the 2nd condition the lim n->inf b_n = 0 is not met, at least that's my approach can someone confirm?

thomas5267 · Answer

If the terms in the series does not approach 0 as it goes to infinity, there is no chance for the series to be convergent.

shamil98 · Answer

oh forgot about this i too the lim of b_n and did l'hopital's 4 times (n^4) and it diverges since its infinity

thomas5267 · Answer

I simply used the fact that exponential function grows faster than all polynomials.

shamil98 · Answer

btw just for clarification since the lim n-> inf = inf
would i say it diverges by the AST or by the divergence test?

thomas5267 · Answer

No idea what it is called. It is called "Term test" on Wikipedia. https://en.wikipedia.org/wiki/Term_test

shamil98 · Answer

yeah the divergence test thats it
alrighty thanks man