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.

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

  • $\begingroup$ But if $a_n <1, \forall n$? $\endgroup$ – GinKin Feb 10 '15 at 16:03
  • $\begingroup$ I see. Why does the given imply that $a_n/n$ is increasing? $\endgroup$ – GinKin Feb 10 '15 at 16:12
  • 1
    $\begingroup$ 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. $\endgroup$ – kobe Feb 10 '15 at 16:14
  • 1
    $\begingroup$ 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. $\endgroup$ – kobe Feb 10 '15 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.