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\lim_{n\rightarrow \infty}\left [ \frac{1}{n} + \frac{n^{2}}{(n+1)^{3}}+ \frac{n^{2}}{(n+2)^{3}} +...+\frac{1}{8n}\right ]=

  • Option 1)

    \frac{3}{8}

  • Option 2)

    0

  • Option 3)

    \frac{1}{4}

  • Option 4)

    \frac{1}{5}

 

Answers (1)

best_answer

 

Walli's Method -

Definite integral by first principle

\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]

where

h=\frac{b-a}{n}

- wherein

 

 \lim_{n\rightarrow \infty } \left [ \frac{1}{n}+\frac{n^{2}}{(n+1)^{3}}+\frac{n^{2}}{(n+2)^{3}}+---+\frac{1}{8n} \right ]

\lim_{n\rightarrow \infty }\left [ \frac{1}{n} +\frac{1}{n(1+\frac{1}{n})^{3}}+\frac{1}{n(1+\frac{2}{n})^{3}}+---+\frac{1}{8n}\right ]

=> \int_{0}^{1}\frac{dx}{(1+x)^{3}}

where \frac{1}{n}=dx

\frac{r}{n}=x

\therefore \int_{0}^{1}\frac{-1}{2}.\frac{1}{(1+x)^{2}}

=-\frac{1}{2}\left [ \frac{1}{4}-\frac{1}{1} \right ]

=\frac{3}{8} 


Option 1)

\frac{3}{8}

Option is Correct

Option 2)

0

Option is incorrect

Option 3)

\frac{1}{4}

Option is incorrect

Option 4)

\frac{1}{5}

Option is incorrect

Posted by

prateek

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