The value of limn - - n-1-n2-n-4-n2-n-9-n2- - 1-2n - is equal to -Bihar CEE 1994-

We have, lim_n →∞ [ n/1+n^2+n/4+n^2+⋯ + 1/2n ] = lim_n →∞∑_r=1^nn/r^2+n^2=lim_n →∞∑_r=1^nn/n^2 ( 1+r^2/n^2 ) = lim_n →∞∑_r=1^n1/n ( 1/1+r^2/n^2 )= ∫_0^1 dx/1+x^ ...

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Standard XI

Mathematics

Definite Integral as Limit of Sum

The value of ...

Question

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]

$\frac{π}{2}$

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$\frac{π}{4}$

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1

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$N o n e o f t h e s e$

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Solution

The correct option is B
$\frac{π}{4}$

We have, ${lim}_{n \to \infty} [\frac{n}{1 + n^{2}} + \frac{n}{4 + n^{2}} + \dots + \frac{1}{2 n}]$
$= {lim}_{n \to \infty} \sum_{r = 1}^{n} \frac{n}{r^{2} + n^{2}} = {lim}_{n \to \infty} \sum_{r = 1}^{n} \frac{n}{n^{2} (1 + \frac{r^{2}}{n^{2}})}$
$= {lim}_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n ⎛ ⎜ ⎜ ⎝ \frac{1}{1 + \frac{r^{2}}{n^{2}}} ⎞ ⎟ ⎟ ⎠} = \int_{0}^{1} \frac{d x}{1 + x^{2}},$
${A p p l y i n g f o r m u l a, {lim}_{n \to \infty} \sum_{r = 1}^{n - 1} {f (\frac{r}{n})} \frac{1}{n} = \int_{0}^{1} f (x) d x}$
$= [t a n^{- 1} x]_{0}^{1} = t a n^{- 1} 1 - t a n^{- 1} 0 = \frac{π}{4}$

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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]

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

is equal to

$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2 n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$ is equal to [RPET 2000]

Q. The value of

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

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The value of limn→∞[n1+n2+n4+n2+n9+n2+⋯+12n] is equal to [Bihar CEE 1994]

The correct option is B π4 We have, limn→∞[n1+n2+n4+n2+⋯+12n] =limn→∞∑nr=1nr2+n2=limn→∞∑nr=1nn2(1+r2n2) =limn→∞∑nr=11n⎛⎜ ⎜⎝11+r2n2⎞⎟ ⎟⎠=∫10dx1+x2, {Applying formula, limn→∞∑n−1r=1{f(rn)}1n=∫10f(x)dx} =[tan−1x]10=tan−11−tan−10=π4

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]