mathop {lim }limits_{n to infty } left[ {frac | vedclass

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(B) Let $y = \mathop {\lim }\limits_{n 	o \infty } \sum_{k=0}^{n-1} \frac{1}{\sqrt{n^2 + kn}}$.We can rewrite the expression as:$y = \mathop {\lim }\

Vedclass · Accepted Answer

$2\sqrt{2} - 2$

Vedclass · Answer

(B) Let $y = \mathop {\lim }\limits_{n 	o \infty } \sum_{k=0}^{n-1} \frac{1}{\sqrt{n^2 + kn}}$.We can rewrite the expression as:$y = \mathop {\lim }\limits_{n 	o \infty } \sum_{k=0}^{n-1} \frac{1}{n\sqrt{1 + \frac{k}{n}}}$.This is a Riemann sum of the form $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{k=0}^{n-1} f\left(\frac{k}{n}ight)$,where $f(x) = \frac{1}{\sqrt{1+x}}$.The limit is equivalent to the definite integral:$y = \int_{0}^{1} \frac{1}{\sqrt{1+x}} dx$.Evaluating the integral:$y = [2\sqrt{1+x}]_{0}^{1}$.$y = 2\sqrt{1+1} - 2\sqrt{1+0}$.$y = 2\sqrt{2} - 2$.

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{\sqrt {{n^2} + n} }} + \frac{1}{{\sqrt {{n^2} + 2n} }} + \dots + \frac{1}{{\sqrt {{n^2} + (n - 1)n} }}} \right]$ is equal to

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