Prove that $a_n=na/(1+n^2 b)$ is strictly decreasing given a, b >0
I have tried proving this directly, by setting
$\frac{na}{1+n^2b} > \frac{(n+1)a}{1+(n+1)^2b}$
and end up with
$\frac{na+n^3ab+2n^2ab+nab}{(1+n^b)(1+n^2b+2bn+b)} > \frac{na+a+n^3ab+n^2ab}{(1+n^b)(1+n^2b+2bn+b)}$
I simplify that down to...
$\frac{n^2b+nb}{(1+n^2b)(1+n^2b+2bn+b)} > \frac{1}{(1+n^2b)(1+n^2b+2bn+b)} $
But I have no particular reason to believe that the inequality is true.  Am I making an algebraic error?  Is there a better way to do this?  Is there a way to do this by induction?
Thanks!
 A: The inequality is false when $n$ and $b$ are small enough. For each fixed $b>0$, the sequence is eventually decreasing.  ($a$ has no effect, as you saw, because it is just a positive number multiplied with everything.)  
Let $f(n)=\dfrac{n}{1+n^2b}$  Then $f(n)=\dfrac{1}{\left(\dfrac{1+n^2b}{n}\right)}=\dfrac{1}{\frac{1}{n}+nb}$.  Thus $f(n+1)<f(n)$ if and only if $\dfrac{1}{n+1}+(n+1)b>\dfrac{1}{n}+nb$ if and only if $b>\dfrac{1}{n(n+1)}$.  While $b$ can be smaller than that for some $n$ (for an arbitrary finite number of them), the inequality will hold for sufficiently large $n$.
You can get the same conclusion from your work.  At your last inequality, the denominators are equal and positive, so the inequality reduces to $n^2b+nb>1$, which is equivalent to $b>\dfrac{1}{n(n+1)}$.
A: The sequence can only be shown to be eventually strictly decreasing.  Take for example $b=\frac{1}{4}$ and $a=1$.  Then we have $a_n=\frac{n}{1+\frac{n^2}{4}}$ so that $a_1=\frac{4}{5}$ and $a_2=1$.  As a hint try treating this as a continuous function $f(x)=\frac{ax}{1+bx^2}$, take the derivative, and try to find a point $x_0$ so that the function is strictly decreasing for all $x>x_0$.  Then think about what you can say about your sequence from this information.
A: Just rewrite it as
$$
a_n = \frac{a}{\frac{1}{n} + nb}.
$$
Since $\frac{1}{n} + nb$ is increasing for sufficiently large $n$ and for $b > 0$, it shows $a_n$ must be decreasing for sufficiently large $n$ and for $a > 0$.
To be rigorous we need to say precisely what sufficiently large means and prove it. By sufficiently large, we mean that there is a $B$ such that $n > B$ implies $\frac{1}{n + 1} + (n + 1)b > \frac{1}{n} + nb$. But this is equivalent to $(n + 1)b - nb > \frac{1}{n} - \frac{1}{n + 1} \iff b > \frac{1}{n(n + 1)}$. But this condition is fulfilled if we have the stronger condition $b > \frac{1}{n^2} \implies n > \frac{1}{\sqrt{b}}$. Hence a $B$ that works is $B = \frac{1}{\sqrt{b}}$.
