Track / Stage 6 / 108 of 400 #1108 of 1964

Problem 1108

National olympiad, first round

Algebra Difficulty 6.2 Prove it

4. Consider the sequence of natural numbers $\left(\mathrm{a}_{\mathrm{n}}\right)_{\mathrm{n} \geq 1}$ , where $\mathrm{a}_{1}=2$ and $\mathrm{a}_{\mathrm{n}+1}=\frac{2 n+3-n!a_{n}}{(n+1)!}, n \geq 1$ .
a.) Show that the sequence is convergent.
4 points
b.) Calculate $\lim _{n \rightarrow \infty} n \cdot \ln \left[(n-1) \cdot a_{n}\right]$
3 points

Prof. Bud Adrian, Negreşti-OaŞ

## Mathematics Olympiad

## local stage, 16.02.2013 Class XI Grading Rubric

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

4.) a.) it is shown by induction that $a_{n}=\frac{n+1}{n!}$ 2 points
demonstrating that the sequence is decreasing. 1 point
$\left(a_{n}\right)_{n \geq 1}$ strictly positive, hence bounded below,
Therefore the sequence is convergent
1 point

b.) We substitute $a_{n}$ with $\frac{n+1}{n!}$ . The limit becomes:

$\lim _{n \rightarrow \infty} n \cdot \ln \left[(n-1)!a_{n}\right]=\lim _{n \rightarrow \infty} n \cdot \ln \frac{n+1}{n}=\lim _{n \rightarrow \infty} \ln \left(1+\frac{1}{n}\right)^{n}=\ln e=1 \quad \ldots \ldots \ldots \ldots . \quad 3$ points

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.