The value of mathop {lim }limits_{n to infty | vedclass

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Question

(A) Let $S = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n {\frac{n}{{{r^2} + {n^2}}}}$.We can rewrite the sum as:$S = \mathop {\lim }\

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$\frac{\pi }{4}$

Vedclass · Answer

(A) Let $S = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n {\frac{n}{{{r^2} + {n^2}}}}$.We can rewrite the sum as:$S = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^n {\frac{n}{{{n^2}(1 + \frac{{{r^2}}}{{{n^2}}})}}} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r = 1}^n {\frac{1}{{1 + {{(\frac{r}{n})}^2}}}}$.Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r = 1}^n {f(\frac{r}{n})} = \int_0^1 {f(x)dx}$,where $f(x) = \frac{1}{{1 + {x^2}}}$.Thus,$S = \int_0^1 {\frac{1}{{1 + {x^2}}}dx}$.Evaluating the integral:$S = [	an^{ - 1} x]_0^1 = 	an^{ - 1}(1) - 	an^{ - 1}(0) = \frac{\pi }{4} - 0 = \frac{\pi }{4}$.

The value of $\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{n}{{1 + {n^2}}} + \frac{n}{{4 + {n^2}}} + \frac{n}{{9 + {n^2}}} + .... + \frac{1}{{2n}}} \right]$ is equal to

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