lim _{n rightarrow infty} nleft[frac{1}{3 n^2 | vedclass

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Question

(B) The given expression is $S = \lim _{n \rightarrow \infty} n \sum_{r=1}^n \frac{1}{3 n^2+8 n r+4 r^2}$.Dividing the numerator and denominator by $n

Vedclass · Accepted Answer

$\frac{1}{4} \log \frac{9}{5}$

Vedclass · Answer

(B) The given expression is $S = \lim _{n ightarrow \infty} n \sum_{r=1}^n \frac{1}{3 n^2+8 n r+4 r^2}$.Dividing the numerator and denominator by $n^2$,we get:$S = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{3+8(\frac{r}{n})+4(\frac{r}{n})^2}$.This is a Riemann sum,which can be written as the definite integral:$S = \int_0^1 \frac{dx}{4x^2+8x+3} = \int_0^1 \frac{dx}{(2x+2)^2-1} = \int_0^1 \frac{dx}{4(x+1)^2-1}$.Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \ln |\frac{x-a}{x+a}|$,we have:$S = \frac{1}{4} \int_0^1 \frac{dx}{(x+1)^2-(\frac{1}{2})^2} = \frac{1}{4} 	imes \frac{1}{2(\frac{1}{2})} [\ln |\frac{x+1-1/2}{x+1+1/2}|]_0^1$.$S = \frac{1}{4} [\ln |\frac{x+1/2}{x+3/2}|]_0^1 = \frac{1}{4} [\ln \frac{3/2}{5/2} - \ln \frac{1/2}{3/2}] = \frac{1}{4} [\ln \frac{3}{5} - \ln \frac{1}{3}] = \frac{1}{4} \ln (\frac{3}{5} 	imes 3) = \frac{1}{4} \ln \frac{9}{5}$.

$\lim _{n \rightarrow \infty} n\left[\frac{1}{3 n^2+8 n+4}+\frac{1}{3 n^2+16 n+16}+\ldots+\frac{1}{15 n^2}\right]=$

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