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How to prove/show that the sequence $a_n=\frac{1}{\sqrt{n^2+1}+n}$ is decreasing? My idea:

  • $n^2<(n+1)^2 /+1$
  • $n^2+1<(n+1)^2+1/ \sqrt{}$
  • $\sqrt{n^2+1}<\sqrt{(n+1)^2+1}/+n$
  • $\sqrt{n^2+1}+n<\sqrt{(n+1)^2+1}+n$

And now I'm stuck since if I add 1 to the both sides, I don't know how to move it from the right side without also moving it from the left side.

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I do not understand first line in your idea.. what does $/ +$ mean? – Praphulla Koushik Mar 22 '14 at 14:28
Can't this be proved by induction on $n$? – Andrew Thompson Mar 22 '14 at 14:28
@PraphullaKoushik With that I meant what I'm doing to the both sides of the inequality: add 1 to the both sides, take the square root to the both sides... – Nick Mar 22 '14 at 14:30
@Nick : I guess that is not a general notation... It would be better if you actually write that in body of the question – Praphulla Koushik Mar 22 '14 at 14:33
up vote 4 down vote accepted

It is enough to show that the sequence of denominators, $\sqrt{1+n^2} + n$, is increasing, since if the denominators increase while the numerators stay the same, the fractions decrease.

We want to show $$\color{darkred}{\sqrt{1+n^2}} + \color{darkblue}{n} < \color{darkred}{\sqrt{1+(n+1)^2}} + \color{darkblue}{(n+1)}.$$ It is enough to show $\color{darkblue}{n < n+1}$ and $\color{darkred}{\sqrt{1+n^2} < \sqrt{1+(n+1)^2}}$. The first is immediate. Because $\sqrt x$ is an increasing function of its argument, it is enough to show $$\begin{align}1+n^2 & < 1+(n+1)^2 \\& = 1 + n^2+2n+1\end{align}.$$ Since $n$ is a positive integer, $2n+1$ is positive and we are done.

(This "increasing function" argument is used constantly, and you should make sure you understand it. If $f(x)$ is an increasing function, then $x>y$ implies $f(x) > f(y)$ and vice versa. Then if we want to show $f(x) > f(y)$, it is enough to show $x>y$. In this example, $f(x) = \sqrt x$. We used the same argument in reverse at the beginning of the answer: since $f(x)= \frac 1x$ is a decreasing function, to show $\frac1x<\frac1y$ it suffices to show that $x>y$.)

(Note also that we could have used this same argument to show that $1+n^2 < 1+(n+1)^2$ instead of the algebraic argument I used above. For positive $n$, the function $1+n^2$ is an increasing function of $n$, so to show $1+n^2 < 1+(n+1)^2$ it suffices to show that $n< n+1$, which needs no proof.)

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It's easy to see ( for example we calculate their derivative) that the functions $$x\mapsto x^2+1\quad;\quad x\mapsto \sqrt x\quad;\quad x\mapsto x$$ are increasing on $[0,+\infty)$ hence the function $$x\mapsto \sqrt{x^2+1}+x$$ is also increasing as composition and sum of increasing functions hence the function $$f\colon x\mapsto \frac1{\sqrt{x^2+1}+x}$$ is decreasing on $[0,\infty)$ since it's the inverse of increasing function. Now what we can say about the monotonicity of the sequence $$a_n=f(n)\qquad?$$

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Note that $\frac1{\sqrt{1+n^2}+n}=\sqrt{1+n^2}-n$.

The sequence would be decreasing if $\sqrt{1+n^2}-n>\sqrt{1+{(n+1)}^2}-n-1\iff \sqrt{\color{grey}{0^2} +1^2}+\sqrt{1+n^2}>\sqrt{1+{(n+1)}^2}$ which is the triangle inequality or Schwarz inequality!.

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You have to prove $$\frac{1}{\sqrt{n^2+1}+n}>\frac{1}{\sqrt{(n+1)^2+1}+n+1}$$

i.e., $$\sqrt{n^2+1}+n< \sqrt{(n+1)^2+1}+n+1$$

i.e., $$\sqrt{n^2+1}< \sqrt{(n+1)^2+1}+1$$

i.e., $$n^2+1< (n+1)^2+1+1+2(\sqrt{(n+1)^2+1})$$

i.e., $$n^2< (n+1)^2+1+2(\sqrt{(n+1)^2+1})$$

i.e., $$0 <2n+1+1+2(\sqrt{(n+1)^2+1})$$

i.e., $$0 <n+(\sqrt{(n+1)^2+1})$$

Can you conclude this and fill gaps in between?

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1) The sequence is decreasing if the denominators are increasing

2) $\sqrt{n^2+1}+n$ is increasing if both $\sqrt{n^2+1}$ and $n$ are increasing

3) $n$ is increasing. $\sqrt{n^2+1}>n$ is also increasing. Q.E.D.

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