Calculus II

[Marie1309]

When we use the ratio test, we find the limit:

[MATH]L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|[/MATH]
The ratio test states that:

• if \(L<1\) then the series converges absolutely;
• if \(L<1\) then the series is divergent;
• if \(L=1\) or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

And so, because we require \(L<1\), the interval of convergence is necessarily open, that is, we do not include the end-points.

Thank you so much. But you put L<1 two times

 

[Steven G]

MarkFL meant to write it is divergent if L>1

Recall that a necessary condition for convergence is that a_n-->0 as n--oo
Now if L>1, then a_n+1> a_n and therefore the (positive) terms are getting bigger and hence not approaching 0. So it diverges
That's how I remember the rules!