$(n+1)x_{n+2}-nx_{n+1}-x_n=0$ ,prove the sequence converges

Question

Sequence $x_n$ for which

$$(n+1)x_{n+2}= nx_{n+1}+x_n$$

for every $n\in\mathbb{N}$. Prove that it converges.

Its not decreasing or increasing, i checked with some random initial values.So, i dont know how to proceed with this. Any help?

Should it be, perhaps, $;(n+1)x_{n-2}=nx_{n+1}+x_n\ldots;$ ? — DonAntonio, Jul 26 '13 at 09:50
the parentheses on the right of x are indexes if thats what u mean. The index of the x with n+1 as coefficient is n+2 though, and thx for the edits, i dont know how to put indexes, sorry :c — Plom, Jul 26 '13 at 09:53
@DonAntonio: It makes no difference. A sequence is a function, and you can write the argument either as an argument or as a subscript. — Brian M. Scott, Jul 26 '13 at 09:54
@DonAntonio: check your suggestion; as written, it is wrong. — Ron Gordon, Jul 26 '13 at 10:05
Well @BrianM.Scott, it makes a difference for me wrt clarity. A matter of habit, I guess. — DonAntonio, Jul 26 '13 at 11:28

score 5 · Accepted Answer · answered Jul 26 '13 at 10:01

The sequence may be rewritten as

$$(n+1) y_{n+1} = -y_n$$

where $y_n = x_{n+1}-x_n$. This then becomes

$$y_n = \frac{(-1)^n}{n!} y_0$$

or

$$x_{n+1}-x_n = \frac{(-1)^n}{n!} (x_1-x_0)$$

Clearly, as $n \to \infty$, $x_{n+1}-x_n \to 0$ and the sequence converges.

score 2 · Answer 2 · answered Jul 26 '13 at 09:58

2

Note that $$(x_{n+2}-x_{n+1})=\frac {(x_n-x_{n+1})}{n+1}$$ which should be enough to get you to an answer - the absolute difference between terms reduces very quickly.

answered Jul 26 '13 at 09:58

Mark Bennet

100,194

(+1) Nice observation. In fact, that shows that the seq. is contraction. – Mhenni Benghorbal Jul 26 '13 at 10:10

score 0 · Answer 3 · answered Jul 26 '13 at 10:10

0

This is an example of a Cauchy Sequence, and a sequence is convergent iff it is cauchy.

answered Jul 26 '13 at 10:10

WhizKid

857

$(n+1)x_{n+2}-nx_{n+1}-x_n=0$ ,prove the sequence converges

3 Answers3