I'm having some difficulties with the following problem:

Let $n\in \mathbb N$ and: $$ x_n = \frac{n-3}{\sqrt{n^2+1}} $$ Prove $x_n$ is a monotonic sequence starting from some $n_0$.

I've applied the following test to prove that. Suppose $\frac{x_n}{x_{n+1}} > 1$

$$ \frac{n-3}{\sqrt{n^2+1}}\cdot\frac{\sqrt{(n+1)^2+1}}{n-2} > 1 \iff \\ \iff \frac{n-3}{n-2} >\frac{\sqrt{n^2+1}}{\sqrt{(n+1)^2+1}} $$

Now squaring both sides (this is BTW where i think things get wrong): $$ \left(\frac{n-3}{n-2}\right)^2 >\frac{{n^2+1}}{{(n+1)^2+1}} \iff \\ \iff (n-3)^2((n+1)^2+1)>(n^2 + 1)(n-2)^2 $$

After some algebraic transformations one may get: $$ 3 n^2 - 5n - 7 < 0 $$

Now find $n_0 \ge 1$ starting from which the inequality holds: $$ n_{1,2} = \frac{5 \pm\sqrt{109}}{6} $$

So from this $n \approx2.57 < 3$. So for $n \ge 3$ the inequality holds. But that's not true! From the graph it's crystal clear that $x_n$ is monotonically increasing $\forall n \ge 1$.

Where did things go wrong? I believe it's a consequence of squaring the inequality, but can't see how it went wrong.

  • $\begingroup$ Hint: what happens when $2 \le n < 3$? $\endgroup$ – Rob Arthan Nov 5 '18 at 18:42
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    $\begingroup$ Even ignoring the potential division by zero, the squaring step introduces a $\implies$ (instead of the $\iff$ at all other steps). So nothing is wrong with concluding "IF the quotient is $>1$ and $n\ge 0$, THEN $n\ge 3$" $\endgroup$ – Hagen von Eitzen Nov 5 '18 at 18:45
  • $\begingroup$ x_0=-3.x_1=-2/√2,x_3=0, increasing .For n>3 start proving. $\endgroup$ – Peter Szilas Nov 5 '18 at 18:58

For all $n\geq3$ we obtain: $$x_{n+1}-x_n=\frac{n-2}{\sqrt{n^2+2n+2}}-\frac{n-3}{\sqrt{n^2+1}}=$$ $$=\frac{2(3n^2-5n-7)}{\sqrt{(n^2+2n+2)(n^2+1)}((n-2)\sqrt{n^2+1}+(n-3)\sqrt{n^2+2n+2})}>0,$$ which says that our sequence increases for $n\geq3$.

  • $\begingroup$ Isn't $x_n$ ($a_n$ in your answer) still monotonic for $n \in [1, 2]>$ $\endgroup$ – roman Nov 5 '18 at 19:20
  • $\begingroup$ @roman It was typo. I fixed. You look for the proof that $x$ is monotonic from some $n_0$. No? I have a proof that your sequence decreases for all natural $n$, but it's another problem. $\endgroup$ – Michael Rozenberg Nov 5 '18 at 19:36
  • $\begingroup$ Well, your answer indeed shows that for $n\ge 3$. I'm just wondering why $n = 1,2$ are missed. $\endgroup$ – roman Nov 5 '18 at 19:54
  • $\begingroup$ @roman Because for $n=1$ the denominator is negative. We need another way. But it's another problem already. $\endgroup$ – Michael Rozenberg Nov 5 '18 at 19:56

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