# Proving $\frac n {a_n+2n}$ converges

Let $a_n$ be a sequence such that $\frac {a_{n+1}}{n+1}>\frac {a_n} n, \forall n \in \mathbb N$.

1. Prove that the sequence $b_n=\frac n {a_n+2n}$ converges.

2. Suppose $a_n$ is bounded, what is $\displaystyle \lim _{n\to\infty}\frac n {a_n+2n}$?

My attempt:

(2). $b_n= \frac 1{\frac {a_n} n +2}$, since $a_n$ is bounded then the limit is $\frac 1 2$.

(1). From the given I can tell $a_n$ is monotone increasing: $\frac {a_{n+1}}{a_n}>\frac {n+1} n> 1\to \frac {a_{n+1}}{a_n}>1 \to {a_{n+1}}>{a_n}$.

Now I probably have to show that $b_n$ is bounded, but I don't see how since the basis of the induction fails for $n=1$: $\frac 1 {a_n+2}\overset{?} <10$ since I have no info on $a_n$.

Also this: $b_n= \frac 1{\frac {a_n} n +2}$ has a $\infty/\infty$ when $n\to \infty$ and so is the given form.

• Are the $a_n$ positive? – Julián Aguirre Feb 10 '15 at 16:15
• @JuliánAguirre unknown. – GinKin Feb 10 '15 at 16:47

You need an additional assumption on $a_n$ for the first part to be true. Having $a_1 > 0$ will do. In that case, since $\frac{a_{n+1}}{n+1} > \frac{a_n}{n}$ for all $n$, $a_{n+1} > (n+1)a_1 > a_1 > 0$ for all $n$. So the sequence $(a_n)$ is positive. Since $b_n = \frac{1}{(a_n/n) + 2}$ and $a_n/n$ is increasing by assumption, $b_n$ is decreasing. Since the $b_n$ are positive, it follows from the monotone convergence theorem for sequences that $b_n$ converges.
• But if $a_n <1, \forall n$? – GinKin Feb 10 '15 at 16:03
• I see. Why does the given imply that $a_n/n$ is increasing? – GinKin Feb 10 '15 at 16:12
• A sequence $(x_n)$ is increasing if $x_n \le x_{n+1}$ for all $n \in \Bbb N$. Setting $c_n = a_n/n$, the condition $a_{n+1}/(n+1) > a_n/n$ for all $n$ is the same as $c_{n+1} > c_n$ for all $n$. So indeed $c_n$, or $a_n/n$, is increasing. – kobe Feb 10 '15 at 16:14
• If you say that $a_n < 1$ for all $n$, then $$b_n = \frac{1}{\frac{a_n}{n} + 2} > \frac{1}{\frac{1}{n} + 2} \ge \frac{1}{1 + 2} = \frac{1}{3}$$ for all $n$, showing that $b_n$ is bounded below (by $1/3$). So in that case, you will be able to claim that $b_n$ converges. – kobe Feb 10 '15 at 16:19