Monotonicity of a sequence

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January 2nd 2008, 07:48 PM
gusztav

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.
January 2nd 2008, 07:52 PM
Jhevon

Quote:

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?
January 2nd 2008, 07:55 PM
gusztav

Quote:

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.