lim _{n rightarrow infty}left[frac{1}{n}+frac | vedclass

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(A) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^{2}}$.We can rewrite the expression as $L = \lim _{n \rightar

Vedclass · Accepted Answer

$\frac{1}{2}$

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(A) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^{2}}$.We can rewrite the expression as $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{n} \cdot \frac{1}{(1 + r/n)^{2}}$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(r/n) = \int_{0}^{1} f(x) dx$.Here,$f(x) = \frac{1}{(1+x)^{2}}$.Thus,$L = \int_{0}^{1} \frac{1}{(1+x)^{2}} dx$.Evaluating the integral: $L = \left[ -\frac{1}{1+x} \right]_{0}^{1} = -\left( \frac{1}{2} - 1 \right) = -(-\frac{1}{2}) = \frac{1}{2}$.

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^{2}}+\frac{n}{(n+2)^{2}}+\ldots+\frac{n}{(2 n-1)^{2}}\right]$ is equal to ...... .

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