$s_n= \frac{1-2+3-4+5-6+7+...+(-2n)}{\sqrt{n^2+1}+\sqrt{n^2-1}}$

by Mathaman Topologius   Last Updated August 12, 2018 13:20 PM

What is the limit of this sequence?

$s_n= \frac{1-2+3-4+5-6+7+...+(-2n)}{\sqrt{n^2+1}+\sqrt{n^2-1}}$

I need hint. I don't want the solution. Which idea should I use here.



Answers 2


$$s_n= \frac{(1-2)+(3-4)+(5-6)+...+(2n-1+(-2n))}{\sqrt{n^2+1}+\sqrt{n^2-1}}=\frac{-n}{\sqrt{n^2+1}+\sqrt{n^2-1}}$$

Riemann
Riemann
August 12, 2018 12:45 PM

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

Arnaud Mortier
Arnaud Mortier
August 12, 2018 12:58 PM

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