Find- n- Lt n p n n-1p n-1 - n P n - -

The correct option is C 0 lim_ n→∞ #160; n _ p _ n n+1 _ p _ n+1 n _ p _ n =lim_ n→∞ #160; n! 0! #160; ( n+1 ) ! 0! n! 0! #1 ...

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Standard XI

Mathematics

Standard Logarithm

Find: n→∞ L...

Question

Find: $L t n \to \infty \frac{^{n} p_{n}}{n + 1 p_{n + 1} -^{n} P_{n}} = ?$

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Solution

The correct option is C $0$
${lim}_{n \to \infty} \frac{n_{p_{n}}}{{n + 1}_{p_{n + 1}} - n_{p_{n}}}$
$= {lim}_{n \to \infty} \frac{\frac{n!}{0!}}{\frac{(n + 1)!}{0!} - \frac{n!}{0!}} = {lim}_{n \to \infty} \frac{n!}{(n + 1)! - n!} = {lim}_{n \to \infty} \frac{n!}{n! (n + 1 - 1)} = {lim}_{n \to \infty} 1 / n = 0$

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Similar questions

l i m n \to \infty \frac{^{n} P_{n}}{^{n + 1} {P_{n + 1} -}^{n} P_{n}}

l i m n \to \infty \frac{1^{P} + 2^{P} + 3^{P} + . . . . + n^{P}}{n^{P + 1}}

equals-

If the number of positive integral solutions of x_1 +x_2+x_3=n be denoted by P_n then
$∣ ∣ ∣ ∣ \begin{matrix} P_{n} & P_{n + 1} & P_{n + 2} P_{n + 1} & P_{n + 2} & P_{n + 3} P_{n + 2} & P_{n + 3} & P_{n + 4} \end{matrix} ∣ ∣ ∣ ∣ =$

L t n \to \infty [\frac{n^{1 / 2}}{(n + 3)^{3 / 2}} + \frac{n^{1 / 2}}{(n + 6)^{3 / 2}} + \frac{n^{1 / 2}}{(n + 9)^{3 / 2}} + . . . + \frac{1}{8 n}]

is equal to :

Q. If

S_{n} = \frac{n (n + 1) (n + 2) (n + 3)}{12}

then prove that

L t n \to \infty \sum_{r = 1}^{n} \frac{1}{t_{r}} = \frac{3}{4}

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Find: Ltn→∞npnn+1pn+1−nPn=?

The correct option is C 0limn→∞npnn+1pn+1−npn=limn→∞n!0!(n+1)!0!−n!0!=limn→∞n!(n+1)!−n!=limn→∞n!n!(n+1−1)=limn→∞1/n=0

Find: $L t n \to \infty \frac{^{n} p_{n}}{n + 1 p_{n + 1} -^{n} P_{n}} = ?$