# Give Counterexample: b_(n+1)-b_n --> 0, then b_n --> L

• Sep 21st 2009, 05:42 PM
Skerven
Give Counterexample: b_(n+1)-b_n --> 0, then b_n --> L
given: a_n = b_(n+1) - b_n

[if the limit of a_n = 0, then (b_n) has a limit] = FALSE.

please provide a counterexample. with all my heart i think the converse is true. if have already proven the converse of the converse.

i have no idea where to even begin, since it seems obvious that ... should the difference between consecutive terms approach zero, then the terms approach an agreeable limit...

i.e. i keep coming up with sequences that actually DO converge (e.g. (-1)^n/n ...)
• Sep 21st 2009, 06:53 PM
NonCommAlg
Quote:

Originally Posted by Skerven
given: a_n = b_(n+1) - b_n

[if the limit of a_n = 0, then (b_n) has a limit] = FALSE.

please provide a counterexample. with all my heart i think the converse is true. if have already proven the converse of the converse.

i have no idea where to even begin, since it seems obvious that ... should the difference between consecutive terms approach zero, then the terms approach an agreeable limit...

i.e. i keep coming up with sequences that actually DO converge (e.g. (-1)^n/n ...)

if you're assuming that L is "finite", then a counter-example would be $\displaystyle b_n = \ln n.$ the converse is obviously true.
• Sep 21st 2009, 08:25 PM
redsoxfan325
Another is $\displaystyle b_n=H_n$, where $\displaystyle H_n=\sum_{k=1}^n\frac{1}{k}$.
• Sep 22nd 2009, 01:25 AM
bkarpuz
$\displaystyle b(n):=\sqrt{n}$ for $\displaystyle n\in\mathbb{N}_{0}$.
By the way, any function satisfying $\displaystyle f,f^{\prime}>0>f^{\prime\prime}$ (these imply that $\displaystyle a>0$ tends to $\displaystyle 0$ asymptotically) with $\displaystyle f(\infty)=\infty$ (this implies $\displaystyle b$ diverges to $\displaystyle \infty$) can be your sequence $\displaystyle b$.
Draw such a function on a paper and try to see what I mean...
• Sep 22nd 2009, 04:33 AM
Skerven
Quote:

Originally Posted by bkarpuz
By the way, any function satisfying $\displaystyle f,f^{\prime}>0>f^{\prime\prime}$ (these imply that $\displaystyle a>0$ tends to $\displaystyle 0$ asymptotically) with $\displaystyle f(\infty)=\infty$ (this implies $\displaystyle b$ diverges to $\displaystyle \infty$) can be your sequence $\displaystyle b$.

Yeah, as strange as it seems, I get it. f(x) -> Inf while f'(x) -> 0. Thanks for that generality.
@redsoxfan325: I also thought of that odd series.