# Sequence convergence

I know that the sequences meets the following:
$$(n+1)(a_{n+1}-a_n)=n(a_{n-1}-a_n)$$
I've got the feeling that this sequence is alternating or decreasing, but I was unable to prove it.
Usually I use induction to prove things about such inductive sequence but in this case I don't have real values t o the first terms of the sequence so I don't have an idea how to even begin inductive prove.

Last edited:

tiny-tim
Homework Helper
hi estro!

(try using the X2 icon just above the Reply box )
(n+1)(a_{n+1}-a_n)=n(a_{n+1}-a_n)

are those two brackets meant to be the same?

if (n+1)X = nX, then X = 0

I have found this question in an old exam, so maybe there is some kind of mistake, I'm not sure.
$$(n+1)(a_{n+1}-a_n)=n(a_{n-1}-a_n)$$
[EDIT] Copied question in a wrong way, now fixed.

tiny-tim
Homework Helper
ah!

ok … standard way of solving that is to define bn = an+1 - an wink:

Still can't see a benefit to define $$b_m=a_{n+1}-a_n$$.
I've used to prove convergence of alternating sequences by using Cantor's Lemma in a geometric formulation.
Marked each section $$[a_{n+1},a_n]$$ and then I proved that each next section smaller and ultimately the length of a sections is going to 0.
But I always had the first terms and used induction to prove sequences is bounded.
But here all these tricks don't work as I can't even prove that this sequence is alternating.

Gib Z
Homework Helper
If you let $b_n = a_n - a_{n-1}$ then your relation is transformed into $$\frac{b_{n+1}}{b_n} = \frac{n}{n+1} < 1$$.