Does This Sequence Converge to \(1-e^{-1}\)?

[magimag]

Homework Statement

Let a₁=0, a₂=1, and a_(n+2)=n*a_(n+1)+a_n/n+1

a)Calculate the value of a₆ and a₇
b)Prove that (a_n) converges.
c)Show that lim a_n=1-e^-1 when n goes to infinity.

The Attempt at a Solution

I got the a part and found out that a₆ 19/30 and a₇)91/144

part b)
each subsequent term becomes less in value, and therefore it is converging. How can prove that it's a Cauchy sequence?

part c)
I'm lost here.

Any hints are appreciated :) I'm lost on this one.

 

[RUber]

Homework Statement

Let a₁=0, a₂=1, and a_(n+2)=n*a_(n+1)+a_n/n+1

Did you forget a parenthesis? What you wrote above does not converge.
$a_1=0, a_2 = 1, a_3 = 1*1+0/2 = 1, a_4 = 2*1+1/3 = 7/3, a_5 = 3*7/3+1/4 =29/4...$
or did you mean:
$a_1=0, a_2 = 1, a_3 = (1*1+0)/2 = 1/2, a_4 = (2*1/2+1)/3 = 2/3, a_5 = (3*2/3+1/2)/4 =5/8, a_6 = (4*5/8 +2/3)/5=19/30?$
Since the latter a_6 matches your post, I will assume that is correct.

I would start by trying some of your basic convergence tests...see if you can pick out some patterns.
Note that if $a_{n+2}=\frac{n a_{n+1}+a_n}{n+1}$, then $a_{n+3}=\frac{(n+1) a_{n+2}+a_{n+1}}{n+2}=\frac{(n+1) \left(\frac{n a_{n+1}+a_n}{n+1}\right)+a_{n+1}}{n+2}$.

 

[pasmith]

Homework Statement

Let a₁=0, a₂=1, and a_(n+2)=n*a_(n+1)+a_n/n+1

a)Calculate the value of a₆ and a₇
b)Prove that (a_n) converges.
c)Show that lim a_n=1-e^-1 when n goes to infinity.

The Attempt at a Solution

I got the a part and found out that a₆ 19/30 and a₇)91/144

part b)
each subsequent term becomes less in value, and therefore it is converging.

This is false: you have \min(a_n,a_{n+1}) \leq a_{n+2} \leq \max(a_n,a_{n+1}) so the sequence is neither increasing nor decreasing.

How can prove that it's a Cauchy sequence?

First, show that a_{n+2} - a_{n+1} = \frac{-1}{n+1}(a_{n+1} - a_n) and hence that a_{n+1} - a_n = \frac{-(-1)^n}{n!}. If n \geq 3 then <br /> |a_{n+1} - a_n| = \frac{1}{n!} &lt; \frac{1}{2^{n-1}}.<br />
It is straightforward to show that a sequence b_n which satisfies <br /> |b_{n+1} - b_n| &lt; \frac{1}{2^{n-1}} for all n \geq N_0 \in \mathbb{N} is cauchy.

part c)
I'm lost here.

I suggested that you prove that <br /> a_{k+1} - a_k = -\frac{(-1)^k}{k!}.<br /> You can then find an expression for a_n from <br /> \sum_{k=1}^{n-1} (a_{k+1} - a_k) = -\sum_{k=1}^{n-1} \frac{(-1)^k}{k!}.<br />