TIFR CSE 2010 | Part A | Question: 7

Question

The limit of $\dfrac{10^{n}}{n!}$ as $n \to \infty$ is.

• A. $0$
• B. $1$
• C. $e$
• D. $10$
• E. $\infty$

Comments

I think this question needs a kinda change, because $n!$ is not defined for non integers, so i think it's has to be nth like nth general term of sequence.

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@Lakshmi Narayana404 Either you take $n$ as positive integer or a positive real, answer will remain the same.

In case of positive integers, you get the limit of a sequence as zero and in case of positive reals, limit of a function as zero.

And also you can extend the factorial for positive reals using the Gamma Function.

Just check with your gate calculator whether you get a value for $0.5!$ or not and compare it with $\frac{\sqrt{\pi}}{2}.$  

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Right now i don't know much about gamma functions, but I just remember a glimpse of it as it's extension of factorial function for non negative $x$, so for this question, as limit of function is zero ( $f(n) = a_n$ $ \forall n\geq x_0$} so it implies the limit of sequence also zero.

Answer 1

$\displaystyle \lim_{n \to \infty} \dfrac{10^n}{n!}$

$= \dfrac{\overbrace{10 \times 10 \times 10 \times \ldots \times 10 \times 10}^{n \text{ times}}}{1 \times 2 \times 3 \times \dots \times (n-1) \times n}$

$ =\overbrace{\frac{10}{1} \cdot \frac{10}{2} \dots \frac{10}{10}}^{\approx 2755} \cdot \underbrace{\frac{10}{11} \dots \frac{10}{100} \cdot \frac{10}{101}}_{\ll 1} \dots \overbrace{\frac{10}{10000} \cdot \frac{10}{10001}}^{\lll 1} \dots \small \text{ goes forever}$

Now we can see that after the $\frac{10}{10}$ term, all subsequent terms are $<1$, and keep decreasing. As we increase the value of $n$ the product will get close to $0$.

So as $n \to \infty$ $\dfrac{10^n}{n!}\to 0$.

Hence, the answer is option A.

Comments

Alternatively:

$10\times 10\times 10....\times 10\leq 10\times 11\times 12....\times (n-1)$

Thus  $10^{n-2}\leq \frac{(n-1)!}{9!}$

Hence we have:

$$\frac{10^{n}}{n!}\leq \frac{100(n-1)!}{9!(n)!} = \frac{100}{9!\cdot n}$$

So the limit approaches to $0$ as $n\rightarrow \infty$

Actually this limit holds true for any real number instead of 10, that is:

$\lim_{n\rightarrow \infty }\frac{x^{n}}{n!}=0$, $x\epsilon R$

Source: https://math.stackexchange.com/questions/77550/prove-that-lim-limits-n-to-infty-fracxnn-0-x-in-bbb-r

Answer 2

To show that the sequence converge, we need to show that the limit of the nth term when n tends to infinity is a constant. 

Usually we can not find limits of factorials, so we need to rewrite our problem a little differently.

 

From that we can conclude:

If we can show that the left hand side and right hand side of the above inequality have the same limit, then the middle expression will have the same limit. (Squeeze theorem or sandwich theorem)

Note that 

and since 10/11 is less than one we have 

Since the previous two limits equal the same number, then 

Hence your sequence converge.

Ref: http://www.enotes.com/homework-help/show-that-sequence-10-n-n-converges-find-its-limit-360310#

Answer 3

Let f(n) = n! and g(n) = 10ⁿ

as n approaches infinity, f(n) grows faster than g(n) as f(n) is asymptotically larger than g(n).

So as n approaches infinity, [g(n) / f(n)] will approach 0 ...

Answer is 0 

Answer 4

i think it is 0  because

  10ⁿ/n!

= 10ⁿ/(nⁿ(1*(1-1/n)*(1-2/n)-----------------------*(1/n)))

=((10/n)*(10/n)................(10/n))*(1/(1*(1-1/n)*(1-2/n)-----------------------*(1/n)))

=0*1/1=0