Originally Posted by

**gusztav** Hi everyone,

I need to determine whether the sequence $\displaystyle (a_n)$, where $\displaystyle 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 $\displaystyle a_n \leq a_{n+1}$ holds for every $\displaystyle n \in \mathbb{N} $. In other words, if the inequality $\displaystyle \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 $\displaystyle a_n \leq a_{n+1}$, and if it *isn't*, then $\displaystyle a_n > a_{n+1}$ 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 $\displaystyle (b_n)$, where $\displaystyle b_n=\frac{2n+5}{5-7n}$, is decreasing or increasing.

We can start by assuming that $\displaystyle (b_n)$ is increasing; in that case $\displaystyle b_n \leq b_{n+1}$ is true if and only if $\displaystyle \frac{2n+5}{5-7n} \leq \frac{2(n+1)+5}{5-7(n+1)} \Leftrightarrow ... \Leftrightarrow -14n^2-39n-10 \leq -14n^2-39n+35 \Leftrightarrow -10 \leq 35$, which *is* true, and therefore the starting assumption ($\displaystyle b_n \leq b_{n+1}$) is also true, and $\displaystyle (b_n)$ is an increasing sequence.

So, basically, I should somehow arrive from this expression: $\displaystyle \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. $\displaystyle a_n \leq a_{n+1}$) could be inferred.

Many thanks.