Letx n -1-1-321-1-621-1-102-1-1- n n -1-22- n - 2Then the value of lim n - xn isA- 1-39C- 1-81D- 0 zero

The correct option is B 19We have nbsp;x_nx_n 1 = ( 1 2n(n+1) )^2 ⇒x_nx_n 1 = ( n+2/n )^2 ( n+1/n 1 )^2 On comparing with the given series, nbsp; x_n = ...

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Standard XII

Mathematics

nth Derivative Test

Letx n =1-1/3...

Question

Let $x_{n} = {(1 - \frac{1}{3})}^{2} {(1 - \frac{1}{6})}^{2} {(1 - \frac{1}{10})}^{2} . . . . {⎛ ⎜ ⎜ ⎜ ⎝ 1 - \frac{1}{\frac{n (n + 1)}{2}} ⎞ ⎟ ⎟ ⎟ ⎠}^{2}, n \geq 2.$
Then the value of $lim n \to \infty x_{n}$ is

\frac{1}{3}

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\frac{1}{9}

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\frac{1}{81}

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0

(zero)

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Solution

The correct option is B $\frac{1}{9}$
We have
$\frac{x_{n}}{x_{n - 1}} = {(1 - \frac{2}{n (n + 1)})}^{2} \Rightarrow \frac{x_{n}}{x_{n - 1}} = \frac{{(\frac{n + 2}{n})}^{2}}{{(\frac{n + 1}{n - 1})}^{2}}$
On comparing with the given series,
$x_{n} = R {(\frac{n + 2}{n})}^{2}$
Here, $R$ is a real constant.

$lim n \to \infty R {(\frac{n + 2}{n})}^{2} = lim n \to \infty R {(1 + \frac{2}{n})}^{2} = A$
Now, $x_{2} = \frac{4}{9}$ thus, $R = \frac{1}{9}$
$∴ lim n \to x x_{n} = lim n \to \infty R {(1 + \frac{2}{n})}^{2} = A = \frac{1}{9}$

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Similar questions

Q. Let

x_{n} = {(1 - \frac{1}{3})}^{2} {(1 - \frac{1}{6})}^{2} {(1 - \frac{1}{10})}^{2} . . . . . . . . {⎛ ⎜ ⎜ ⎜ ⎝ 1 - \frac{1}{\frac{n (n + 1)}{2}} ⎞ ⎟ ⎟ ⎟ ⎠}^{2}, n \geq 2

.
Then the value of

lim n \to \infty x_{n}

Q. Let

x_{n} = {(1 - \frac{1}{3})}^{2} {(1 - \frac{1}{6})}^{2} {(1 - \frac{1}{10})}^{2} . . . {⎛ ⎜ ⎝ 1 - \frac{1}{\frac{n (n + 1)}{2}} ⎞ ⎟ ⎠}^{2}, n \geq 2

. Then, the calculate of

lim n \to \infty X_{n}

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

equals

Q. The value of

\lim_{n \to \infty} \{\frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2}\}

is

(a) 1/2

(b) 1

(c) −1

(d) −1/2

Q. Let

P_{n} = {(1 - \frac{1}{3})}^{2} {(1 - \frac{1}{6})}^{2} \dots {(1 - \frac{2}{n (n + 1)})}^{2},

where

n \geq 2 (n \in N) .

lim n \to \infty P_{n} = \frac{a}{b},

where

a

and

b

are coprime, then the value of

a + b

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Letxn=(1−13)2(1−16)2(1−110)2....⎛⎜ ⎜ ⎜⎝1−1n(n+1)2⎞⎟ ⎟ ⎟⎠2, n≥2. Then the value of limn→∞xn is

The correct option is B 19We have xnxn−1=(1−2n(n+1))2⇒xnxn−1=(n+2n)2(n+1n−1)2 On comparing with the given series, xn=R(n+2n)2 Here, R is a real constant. limn→∞R(n+2n)2=limn→∞R(1+2n)2=A Now, x2=49 thus, R=19 ∴limn→xxn=limn→∞R(1+2n)2=A=19

Let $x_{n} = {(1 - \frac{1}{3})}^{2} {(1 - \frac{1}{6})}^{2} {(1 - \frac{1}{10})}^{2} . . . . {⎛ ⎜ ⎜ ⎜ ⎝ 1 - \frac{1}{\frac{n (n + 1)}{2}} ⎞ ⎟ ⎟ ⎟ ⎠}^{2}, n \geq 2.$
Then the value of $lim n \to \infty x_{n}$ is