Let a_n = frac{10^n}{n!} for n = 1, 2, 3, ldo | vedclass

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(C) To find the value of $n$ for which $a_n = \frac{10^n}{n!}$ is maximum,we examine the ratio $\frac{a_n}{a_{n-1}}$.$\frac{a_n}{a_{n-1}} = \frac{10^n

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$10$

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(C) To find the value of $n$ for which $a_n = \frac{10^n}{n!}$ is maximum,we examine the ratio $\frac{a_n}{a_{n-1}}$.$\frac{a_n}{a_{n-1}} = \frac{10^n}{n!} 	imes \frac{(n-1)!}{10^{n-1}} = \frac{10}{n}$.For $a_n$ to be increasing,we require $\frac{a_n}{a_{n-1}} > 1$,which implies $\frac{10}{n} > 1$,so $n This means $a_1 For $n = 10$,$\frac{a_{10}}{a_9} = \frac{10}{10} = 1$,which implies $a_{10} = a_9$.For $n > 10$,$\frac{a_n}{a_{n-1}} Thus,the sequence $a_n$ reaches its maximum value at both $n = 9$ and $n = 10$.Since $10$ is the greatest such value,the answer is $10$.

Let $a_n = \frac{10^n}{n!}$ for $n = 1, 2, 3, \ldots$. The greatest value of $n$ for which $a_n$ is the greatest is:

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