1Actually the answer's zero.
Here goes my first question in TargetIIT :):):):):):)
Find the following limits -
1 > lim [ ( n ) !n n ]
2 > lim [ ( n - 1 ) !n n - 1.5 ] e n
In both cases n tends to infinity .
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21 Answers
1STIRLINGS APPROXIMATION -- pretty interesting thing though :o :o !!
1WOWWWW !!!!!! How did Gallardo do the generalisation , i mean the question ?????? When I fist saw the qs, , I was left completely dumbstruck !!!
Gallardo bhai , KUDOSSSSS
1I think
1-(n-1)n
is not equal to ∞ as n →∞
since, 1-∞∞ = 1-1 =0
is this correct
1Rearranging , the givrn term =n ! 2007 ! n2007( n + 2007 ) !
={ nn n2007 } { 2007 ! } { √2 pi n } { e - n }( n + 2007 ){ n + 2007 } { e - n - 2007 } { √ 2 pi ( n + 2007 ) }
={ 2007 ! } { n n + 2007 } { .......}{ ...... } = { 2007 ! } { [ n + 2007 ] / n } n + 2007 { e - n - 2007 + n } √[ n + 2007 ] / n
={ 2007 ! }{ 1 + [ 2007 / n ] } n + 2007 { e - 2007 } { √ 1 + [ 2007 / n ] }
= 2007 !e - 2007 e 2007 = 2007 ! , after putting the limit .
1@gallardo
first of all r u a student....?
ur ways doesnt seem ur r student becuz if i know correctly stirling approximation is neither in any olympiad topic and obviously not in jee.....so how u know abt it...?
btw how will u evaluate this using stirling approx. ;)
\lim_{n\to\infty}\;\;\;\left(\frac{n!\;\cdot\;n^{2007}}{2008\cdot2009\cdot2010\cdot2011\cdot2012\cdot\;\;\;\cdots\;\;\;\cdot (2007+n)}\right)
goodluck :)
24that approx has been used a lot of times here..[1]
so dont worry [6]
1OKKK , here is the solution ---
There is something called STIRLING'S APPROXIMATION , which states that for large enough n , we are allowed to write
n ! = n n e - n √2 Î n
or , ( n - 1 ) ! x n n n e n = √2 Î x n . 5
or , ( n - 1 ) !n n - 1 . 5 e n = √2 Î
or , k = √2 Î ( suppose )
Hence , limn--> ∞ k = √2 Î
Actually , the expression that I have written for stirlings approximation is approximately correct , and the real expression is pretty big and involves big - O notation of Gauss and many other complicated things which is again very tough to grasp at once - but the result won't differ anyway . If you want to know more , just use WIKIPEDIA
49checking my basics!! lim (ex-1)/x = 1 right?? or is this wrng???i mean is it ex-1 or ex-1???
x->0
492nd one apparently is either 0 or infinite!! not got any strong proof yet tho!![2][2]
29Ans 1 \frac{n!}{n^{n}} = \frac{n(n-1)(n-2)...1}{n^{n}} = \frac{n^{n}{(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})...}}{n^{n}} = 0..at.. x - infinty
49\frac{n!}{n^{n}}=\frac{n(n-1)(n-2)(n-3)(n-4)....3.2.1}{n.n.n.n.n.....n.n.n}=\frac{n}{n}.\frac{n-1}{n}.\frac{n-2}{n}......\frac{3}{n}.\frac{2}{n}.\frac{1}{n}
lim n->∞ => 1/n->0...hence the result!![1][1]