11wow....gr8 solution...
Thanks Pritish..
What is
\huge \lim_{n \rightarrow \propto }\frac{x^{n}}{n!}
?????
ans :0
(Right now I am studying in a state engg colg. in 1st yr. I came across this qsn once in success magnet of Aakash Inst. and another time while the teacher was solving a problem of infinite series. On asking the teacher to show how the limit is solved he told I can show it but you won't understand..it involves higher math concept..you will learn it later...(With due respect to the teacher)...Does it really involves some high level maths??? some one please help how to sove this...
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10 Answers
39Let the limit be some value L. Take logarithm on both sides,
Limn→∞ [nlog(x) - log(n!)] = logL
This is the natural logarithm. For log(n!), we use Stirling's formula -:
log(n!) = nlog(n) - n + more terms which can be neglected.
Limn→∞ [nlog(x) - nlog(n) + n] = logL
Now you can see that the LHS will always be infinity whatever be the value of x.
logL = ∞, this happens when L = 0.
Hence the limit is 0.
Maybe his higher level concept involved gamma functions...
341There are two ways to look at it
1) By the archimedean property of real numbers there exists a natural number N such that N>x (i took the liberty of using this concept as u said u r in engg. otherwise you say its just plain common sense :D)
Let K=\frac{x^N}{N!}
and let L = \frac{x}{N+1}<1
Then
0< \frac{x^n}{n!} < K L^{n-N} = \left(\frac{K}{L^N}\right) L^n
and since \left(\frac{K}{L^N}\right)\lim_{n \rightarrow \infty} L^n = 0 by sandwich theorem the required limit is zero.
2) We know that if the infinite series tex]\sum_{n=1}^{\infty}a_n is convergent then \lim_{n\rightarrow \infty}a_n=0
This follows from the Cauchy criterion on the series a_1, a_1+a_2, a_1+a_2+a_3,..., a_1+a_2+...+a_n,...
Since e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} is convergent we must have \lim_{n \rightarrow \infty} \frac{x^n}{n!} = 0
11sir i did not understand how this came..
0< \frac{x^n}{n!} < K L^{n-N} = \left(\frac{K}{L^N}\right) L^n....
sir i cud'nt understand how are we getting xnn! < K Ln-N ???
341\frac{x^n}{n!} = \frac{x^N}{N!} \times \frac{x}{N+1} \times \frac{x}{N+2}\times...\times \frac{x^n}{n}
Now its obvious that \frac{x}{N+k} < \frac{x}{N+1} for k>1.
Hence
RHS<K \left(\frac{x}{N+1}\right)^{n-N} = KL^{n-N}
39There was bound to be something wrong with my solution :P
21somebody check if my solution is correct....
L=lim n→∞ (xn / n! )
log L =nlog x - log n!= nlog x-log(n/e)n - log√(2 pi n)
=nlog(xe/n) - log √(2 pi n)
so log L = -∞ L=0
341That's a clean solution using Stirling Approximation for n!
The proof in my post gives a sense of numerically why it should go to zero. Also, if you write out the sequence, it goes on increasing for a while and then begins to drop off. So its a good illustration that for the limit of a sequence, the behaviour up to a countable (perhaps infinite) number of elements may be ignored - its the behaviour that takes place eventually that counts.
If you guys have the time, it may be worth your while to read about Goodstein sequences that all counter-intuitively converge to zero. The first few terms rapidly increase to some astronomical magnitude and then astonishingly decrease to zero. Isnt that fantastic!