NLt- -1-1-n2-2-1-n2-n-1-n2-

The correct option is C 1/2 nbsp;lim _ n→∞ nbsp;( 1 / 1 n ^ 2 + 2 / 1 n ^ 2 +... n / 1 n ^ 2 nbsp;) nbsp; nbsp;=lim _ n→∞ nbsp;( 1+2 ...

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Definite Integral as Limit of Sum

nLt→∞ [1/1-n2...

Question

$n L t \to \infty [\frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + \dots + \frac{n}{1 - n^{2}}] =$

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Similar questions

n L t \to \infty [\frac{n + 1}{n^{2} + 1^{2}} + \frac{n + 2}{n^{2} + 2^{2}} + \dots + \frac{n + n}{n^{2} + n^{2}}] =

n L t \to \infty {\frac{1}{\sqrt{n^{2} + 1^{2}}} + \frac{1}{\sqrt{n^{2} + 2^{2}}} + \dots + \frac{1}{\sqrt{n^{2} + n^{2}}}} =

n L t \to \infty [\frac{1}{n} + \frac{n^{2}}{(n + 1)^{3}} + \frac{n^{2}}{(n + 2)^{3}} + \dots . + \frac{1}{8 n}] =

lim n \to \infty (\frac{1}{n^{2} + 1} + \frac{1}{n^{2} + 2} + \frac{1}{n^{2} + 3} + . . . + \frac{n}{n^{2} + n})

lim x \to \infty (\frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + \frac{3}{1 - n^{2}} + . . . + \frac{n}{1 - n^{2}})

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