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I need to prove the following:

$\sqrt{n^2 + 1} - n$ is decreasing

In other words, prove: $\sqrt{(n+1)^2 + 1} - (n+1) < \sqrt{n^2 + 1} - n$

Logically I understand why this holds, since the bigger $n$ gets, the closer $\sqrt{n^2 + 1}$ gets to $\sqrt{n^2} = n$, but I don't know how to prove this algebraically.

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3 Answers 3

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hint

Multiply by the conjugate to get

$$\sqrt {n^2+1}-n=\frac {1}{\sqrt {n^2+1}+n} $$

the sequence in the denominator is increasing.

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  • $\begingroup$ Best way of course! (+1) $\endgroup$
    – user
    Mar 28, 2018 at 22:15
  • $\begingroup$ (+1) Your way is better than mine. $\endgroup$ Mar 28, 2018 at 22:25
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Thinking of $n$ as a real variable we find that $$\frac{\mathrm{d}}{\mathrm{d}n} \left( \sqrt{n^2+1}-n \right)=\frac{n}{\sqrt{n^2+1}}-1<0. $$

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You have\begin{multline}\sqrt{(n+1)^2+1}-(n+1)<\sqrt{n^2+1}-n\iff\\\iff\sqrt{(n+1)^2+1}-\sqrt{n^2+1}<n+1-n=1.\end{multline}But$$\sqrt{(n+1)^2+1}-\sqrt{n^2+1}=\frac{2n+1}{\sqrt{(n+1)^2+1}+\sqrt{n^2+1}}$$and it is true that this is less than $1$, because$$\sqrt{(n+1)^2+1}+\sqrt{n^2+1}>\sqrt{(n+1)^2}+\sqrt{n^2}=2n+1.$$

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