limn→∞[1n+n(n+1)2+n(n+2)2+......+n(2n−1)2] is equal to

limn→∞∑r=0n−1n (n+r)2 limn→∞∑r=0n−11n.1 (1+rn)2=∫011 (1+x)2= [−11+x]01=−12− (−1)=12

$\underset{n \to \infty}{l i m} [\frac{1}{n} + \frac{n}{{(n + 1)}^{2}} + \frac{n}{{(n + 2)}^{2}} + . . . . . . + \frac{n}{{(2 n - 1)}^{2}}]$ is equal to

Option 1 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>

Option 3 - 1

Option 4 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>

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Payal Gupta | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

$\underset{n \to \infty}{l i m} \sum_{r = 0}^{n - 1} \frac{n}{{(n + r)}^{2}}$

$\underset{n \to \infty}{l i m} \sum_{r = 0}^{n - 1} \frac{1}{n} . \frac{1}{{(1 + \frac{r}{n})}^{2}} = \int_{0}^{1} \frac{1}{{(1 + x)}^{2}} = {[- \frac{1}{1 + x}]}_{0}^{1} = - \frac{1}{2} - (- 1) = \frac{1}{2}$

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