mathop {lim }limits_{n to infty } sumlimits_{ | vedclass

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Question

(A) Let $I = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $. Divide the numerator and denominator by $n^2$

Vedclass · Accepted Answer

$\frac{1}{2}\log 2$

Vedclass · Answer

(A) Let $I = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $. Divide the numerator and denominator by $n^2$: $I = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{k = 1}^n {\frac{k/n^2}{1 + (k/n)^2}} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{k = 1}^n {\frac{k/n}{1 + (k/n)^2}}$. Using the definition of a definite integral as the limit of a sum,where $\frac{k}{n} = x$ and $\frac{1}{n} = dx$,as $n 	o \infty$,the sum becomes: $I = \int\limits_0^1 {\frac{x}{{1 + {x^2}}}dx} $. Let $u = 1 + x^2$,then $du = 2x dx$,or $x dx = \frac{1}{2} du$. When $x = 0, u = 1$. When $x = 1, u = 2$. $I = \frac{1}{2} \int\limits_1^2 {\frac{1}{u} du} = \frac{1}{2} [\log |u|]_1^2 = \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2$.

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $ is equal to

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