Lim n - n - n 2-12- n - n 2-22- n - n 2-32-1-5 n is equal to -A- - - 4 B- tan -12C- tan -13D- - - 2

Lim_n→∞ ( nn^2+1^2+nn^2+2^2+nn^2+3^2+...+15n ) =lim_n→∞ ( nn^2+1^2+nn^2+2^2+nn^2+3^2+...+nn^2+(2n)^2 ) =lim_n→∞∑_r=1^2nnn^2+r^2 =lim_n→∞∑_r=1^2n1n ( 11+ ( rn )^ ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

lim n →∞ n / ...

Question

$lim n \to \infty (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + . . . + \frac{1}{5 n})$

is equal to :

{tan}^{- 1} (2)

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{tan}^{- 1} (3)

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π / 4

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Solution

The correct option is A ${tan}^{- 1} (2)$
$lim n \to \infty (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + . . . + \frac{1}{5 n})$
$= lim n \to \infty (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + . . . + \frac{n}{n^{2} + (2 n)^{2}})$
$= lim n \to \infty 2 n \sum r = 1 \frac{n}{n^{2} + r^{2}}$
$= lim n \to \infty 2 n \sum r = 1 \frac{1}{n} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{1}{1 + {(\frac{r}{n})}^{2}} ⎞ ⎟ ⎟ ⎟ ⎠$
$= 2 \int 0 \frac{d x}{1 + x^{2}} = {tan}^{- 1} (2)$

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Similar questions

lim n \to \infty (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + . . . + \frac{1}{5 n})

is equal to :

lim n \to \infty (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} . . . . . . \frac{1}{5 n})

The value of ${lim}_{n \to \infty} [\frac{n}{1 + n^{2}} + \frac{n}{4 + n^{2}} + \frac{n}{9 + n^{2}} + \dots + \frac{1}{2 n}]$ is equal to [Bihar CEE 1994]

Q. What is

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

equal to?

{lim}_{n \to \infty} ⎡ ⎢ ⎢ ⎣ \frac{1}{n} + \frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + . . . . + \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}} ⎤ ⎥ ⎥ ⎦

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limn→∞(nn2+12+nn2+22+nn2+32+...+15n) is equal to :

The correct option is A tan−1(2)limn→∞(nn2+12+nn2+22+nn2+32+...+15n) =limn→∞(nn2+12+nn2+22+nn2+32+...+nn2+(2n)2) =limn→∞2n∑r=1nn2+r2 =limn→∞2n∑r=11n⎛⎜ ⎜ ⎜⎝11+(rn)2⎞⎟ ⎟ ⎟⎠ =2∫0dx1+x2=tan−1(2)

$lim n \to \infty (\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \frac{n}{n^{2} + 3^{2}} + . . . + \frac{1}{5 n})$

is equal to :