JEE Main & Advanced Mathematics Definite Integration Question Bank Summation of series by Definite Integration, Gamma function, Leibnitz's rule

  • question_answer
    \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{2n}{\frac{r}{\sqrt{{{n}^{2}}+{{r}^{2}}}}}\] equals   [IIT 1997 Re-exam]

    A)                 \[1+\sqrt{5}\]   

    B)                 \[-1+\sqrt{5}\]

    C)                 \[-1+\sqrt{2}\] 

    D)                 \[1+\sqrt{2}\]

    Correct Answer: B

    Solution :

                    \[L=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\sum\limits_{r=1}^{2n}{{}}\frac{1}{n}\,.\,\frac{r/n}{\sqrt{1+{{(r/n)}^{2}}}}\]=\[\int_{0}^{2}{{}}\frac{x}{\sqrt{1+{{x}^{2}}}}\,dx=\sqrt{5}-1\].


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