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

1. ## 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 ...)

2. 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 $b_n = \ln n.$ the converse is obviously true.

3. Another is $b_n=H_n$, where $H_n=\sum_{k=1}^n\frac{1}{k}$.

4. $b(n):=\sqrt{n}$ for $n\in\mathbb{N}_{0}$.
By the way, any function satisfying $f,f^{\prime}>0>f^{\prime\prime}$ (these imply that $a>0$ tends to $0$ asymptotically) with $f(\infty)=\infty$ (this implies $b$ diverges to $\infty$) can be your sequence $b$.
Draw such a function on a paper and try to see what I mean...

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