Question

$n\stackrel{Lt}{\to }\infty \{\frac{1}{\sqrt{n^2+1^2}}+\frac{1}{\sqrt{n^2+2^2}}+\dots +\frac{1}{\sqrt{n^2+n^2}}\}=$

• A. $\log (\sqrt{2}+2)$
• B. $\log (\sqrt{2}-2)$
• C. $\log (\sqrt{2}+1)$
• D. $\log (\sqrt{2}-1)$

Answer 1

The correct option is C $\log (\sqrt{2}+1)$

$l{t}_{n\to \infty }\frac{1}{n}[\frac{1}{\sqrt{1+{(}^1{/}_n{)}^2}}+\frac{1}{\sqrt{1+{(}^2{/}_n{)}^2}}+.........]$
$l{t}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^n\frac{1}{\sqrt{1+{(}^r{/}_n{)}^2}}$
$={\int }_0^1\frac{1}{\sqrt{1+x^2}}dx$
$=log|x+\sqrt{x^2+1}|{|}_0^1$
$=log|1+\sqrt{2}|$
$=log(\sqrt{2}+1)$