lim _{n rightarrow infty}left(frac{n^{2}}{lef | vedclass

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Question

(A) The given expression can be written as $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{n^{2}}{(n^{2}+r^{2})(n+r)}$.Dividing the numerator and d

Vedclass · Accepted Answer

$\frac{\pi}{8}+\frac{1}{4} \log _{ e } 2$

Vedclass · Answer

(A) The given expression can be written as $\lim _{n ightarrow \infty} \sum_{r=1}^{n} \frac{n^{2}}{(n^{2}+r^{2})(n+r)}$.Dividing the numerator and denominator by $n^3$,we get:$\lim _{n ightarrow \infty} \sum_{r=1}^{n} \frac{1}{n \left(1+(\frac{r}{n})^{2}ight)(1+\frac{r}{n})}$.This is a Riemann sum,which can be expressed as the definite integral:$\int_{0}^{1} \frac{dx}{(1+x^{2})(1+x)}$.Using partial fractions,$\frac{1}{(1+x^{2})(1+x)} = \frac{A}{1+x} + \frac{Bx+C}{1+x^{2}}$. Solving gives $A = \frac{1}{2}$,$B = -\frac{1}{2}$,$C = \frac{1}{2}$.So,the integral becomes $\frac{1}{2} \int_{0}^{1} \frac{dx}{1+x} - \frac{1}{2} \int_{0}^{1} \frac{x-1}{1+x^{2}} dx$.$= \frac{1}{2} [\ln(1+x)]_{0}^{1} - \frac{1}{2} [\frac{1}{2} \ln(1+x^{2}) - 	an^{-1} x]_{0}^{1}$.$= \frac{1}{2} \ln 2 - \frac{1}{2} (\frac{1}{2} \ln 2 - \frac{\pi}{4}) = \frac{1}{2} \ln 2 - \frac{1}{4} \ln 2 + \frac{\pi}{8} = \frac{\pi}{8} + \frac{1}{4} \ln 2$.

$\lim _{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ is equal to

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