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Let a_{n}= product of the first n natural numbers. Then for al

Question

Let $a_{n}$ = product of the first n natural numbers. Then for all $n ϵ N$

n^{n} \geq a_{n}

(\frac{n + 1}{2})^{n} \geq n!

n^{n} \geq a_{n + 1}

none of these

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Eleanor Discussed

Let

a_{n}

= product of the first n natural numbers. Then for all

n ϵ N

5 mins ago

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Text solutionVerified

Given that,

n ϵ N

\Rightarrow n \geq 1

Therefore,

n \geq \frac{n + 1}{2}

...(1)

a_{n} = 1.2.3..... n = n!

Since,

A . M \geq G . M

Therefore,

\frac{1 + 2 + 3... + n}{n} \geq (1.2.3..... n)^{1/ n}

\Rightarrow \frac{n ( n + 1 )}{2 n} \geq (n!)^{1/ n}

\Rightarrow (\frac{n + 1}{2})^{n} \geq n!

and From (1)

(n)^{n} \geq n!

Ans: A,B

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Question Text	Let $a_{n}$ = product of the first n natural numbers. Then for all $n ϵ N$
Answer Type	Text solution:1
Upvotes	150