# Math Help - Monotonicity of a sequence

1. ## Monotonicity of a sequence

Hi everyone,
I need to determine whether the sequence $(a_n)$, where $a_n=\frac{
\sqrt{n^2+1}-n}
{\sqrt{n+3}+2n}$
, is decreasing or increasing. To do that, I can first suppose that it's e.g. increasing, and then see if the relation $a_n \leq a_{n+1}$ holds for every $n \in \mathbb{N}$. In other words, if the inequality $\frac{
\sqrt{n^2+1}-n}
{\sqrt{n+3}+2n}\leq
\frac{\sqrt{(n+1)^2+1}-(n+1)}
{\sqrt{(n+1)+3}+2(n+1)}$
is true, then is also $a_n \leq a_{n+1}$, and if it isn't, then $a_n > a_{n+1}$ and the sequence is decreasing. This is where I got stuck, for no matter how I tried to solve this inequality (squaring both sides; multiplying whole inequality by both denominators etc), I can't seem to arrive at the solution.

To give you a better idea of the way I am supposed to solve this, here is an easier example:
determine whether the sequence $(b_n)$, where $b_n=\frac{2n+5}{5-7n}$, is decreasing or increasing.
We can start by assuming that $(b_n)$ is increasing; in that case $b_n \leq b_{n+1}$ is true if and only if $\frac{2n+5}{5-7n} \leq \frac{2(n+1)+5}{5-7(n+1)} \Leftrightarrow ... \Leftrightarrow -14n^2-39n-10 \leq -14n^2-39n+35 \Leftrightarrow -10 \leq 35$, which is true, and therefore the starting assumption ( $b_n \leq b_{n+1}$) is also true, and $(b_n)$ is an increasing sequence.

So, basically, I should somehow arrive from this expression: $\frac{
\sqrt{n^2+1}-n}
{\sqrt{n+3}+2n}\leq
\frac{\sqrt{(n+1)^2+1}-(n+1)}
{\sqrt{(n+1)+3}+2(n+1)}$
to a statement from which the verity of the starting assumption (i.e. $a_n \leq a_{n+1}$) could be inferred.

Many thanks.

2. Originally Posted by gusztav
Hi everyone,
I need to determine whether the sequence $(a_n)$, where $a_n=\frac{
\sqrt{n^2+1}-n}
{\sqrt{n+3}+2n}$
, is decreasing or increasing. To do that, I can first suppose that it's e.g. increasing, and then see if the relation $a_n \leq a_{n+1}$ holds for every $n \in \mathbb{N}$. In other words, if the inequality $\frac{
\sqrt{n^2+1}-n}
{\sqrt{n+3}+2n}\leq
\frac{\sqrt{(n+1)^2+1}-(n+1)}
{\sqrt{(n+1)+3}+2(n+1)}$
is true, then is also $a_n \leq a_{n+1}$, and if it isn't, then $a_n > a_{n+1}$ and the sequence is decreasing. This is where I got stuck, for no matter how I tried to solve this inequality (squaring both sides; multiplying whole inequality by both denominators etc), I can't seem to arrive at the solution.

To give you a better idea of the way I am supposed to solve this, here is an easier example:
determine whether the sequence $(b_n)$, where $b_n=\frac{2n+5}{5-7n}$, is decreasing or increasing.
We can start by assuming that $(b_n)$ is increasing; in that case $b_n \leq b_{n+1}$ is true if and only if $\frac{2n+5}{5-7n} \leq \frac{2(n+1)+5}{5-7(n+1)} \Leftrightarrow ... \Leftrightarrow -14n^2-39n-10 \leq -14n^2-39n+35 \Leftrightarrow -10 \leq 35$, which is true, and therefore the starting assumption ( $b_n \leq b_{n+1}$) is also true, and $(b_n)$ is an increasing sequence.

So, basically, I should somehow arrive from this expression: $\frac{
\sqrt{n^2+1}-n}
{\sqrt{n+3}+2n}\leq
\frac{\sqrt{(n+1)^2+1}-(n+1)}
{\sqrt{(n+1)+3}+2(n+1)}$
to a statement from which the verity of the starting assumption (i.e. $a_n \leq a_{n+1}$) could be inferred.

Many thanks.
hmm... i'd use induction (though it would probably be messy here). find $a_1$ and $a_2$ and base your assumption on those values (that is, if $a_1 \ge a_2$ then assume it is decreasing, otherwise assume it is increasing). then try to prove the rest by induction. do you know how to do mathematical induction?

3. Originally Posted by Jhevon
hmm... i'd use induction (though it would probably be messy here). find $a_1$ and $a_2$ and base your assumption on those values (that is, if $a_1 \ge a_2$ then assume it is decreasing, otherwise assume it is increasing). then try to prove the rest by induction. do you know how to do mathematical induction?
Yes. It's a great idea, thanks, I'll try it.