Calculate lim n->inf (n^2/n!)

  • Thread starter cloveryeah
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  • #1
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Homework Statement


calculate lim n->inf (n^2/n!)

2. Homework Equations



The Attempt at a Solution


i just calculated, and i got inf as the ans...
n^2/1X2X3...Xn
n/1X2X3...X(n-1)
1/[(1X2X3...)/n](1-1/n)
when n->inf, it will go to inf

am i correct? or there is something wrong?
 

Answers and Replies

  • #2
SammyS
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Homework Statement


calculate lim n->inf (n^2/n!)

2. Homework Equations



The Attempt at a Solution


i just calculated, and i got inf as the ans...
n^2/1X2X3...Xn
n/1X2X3...X(n-1)
1/[(1X2X3...)/n](1-1/n)
when n->inf, it will go to inf

am i correct? or there is something wrong?
That is incorrect.
 
  • #3
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what is wrong?
 
  • #4
35,287
7,138
what is wrong?
Your answer is wrong. The limit is not infinity.

cloveryeah said:
n^2/1X2X3...Xn
n/1X2X3...X(n-1)
1/[(1X2X3...)/n](1-1/n)
The 2nd and 3rd lines above are OK but written ambiguously.
According to the standard rules of precedence, n/1X2X3...X(n-1) = ##\frac n 1 2 \cdot 3 \cdot \dots (n - 1)##, which is probably not what you intended.
The 3rd line would be interpreted as ##\frac 1 { \frac{1 \cdot 2 \cdot 3 \dots } n} (1 - 1/n)##. That's so convoluted that you probably misled yourself.
 
  • #6
35,287
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why not
n / [1X2X3...(n-1)]
That would be fine, although at the calculus level, you should get out of the habit of writing X for multiplication. It's too easily confused as a variable.

So what is ##\lim_{n \to \infty}\frac n {1 \cdot 2 \cdot 3 \cdot \dots (n - 1)}##?
 
  • #7
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lim→∞ 1/ [ (1⋅2⋅3⋅⋅⋅/n)(1-1/n)] = 1/0 = inf ??
 
  • #8
Ray Vickson
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lim→∞ 1/ [ (1⋅2⋅3⋅⋅⋅/n)(1-1/n)] = 1/0 = inf ??

You tell us.
 
  • #9
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@cloveryeah, here are some easier examples for you, to help you get your head around this stuff.

##\frac {3^2} {3!} = \frac 9 6 = 3/2##
##\frac {4^2} {4!} = \frac {16}{24} = 2/3##
##\frac {5^2} {5!} = ?##
##\frac {6^2} {6!} = ?##
Do you see any sort of trend here?
 
  • #10
pasmith
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Consider: If [itex]n > 4[/itex] then [itex]n! = n(n-1) \times \cdots \times 1 > n(n-1)[/itex]. Thus [itex]0 < \frac{n^2}{n!} < \frac{n^2}{n(n-1)} < \frac43[/itex]. That tells you that the limit of [itex]\frac{n^2}{n!}[/itex], if it exists, must be finite. Unfortunately this bound on [itex]\frac1{n!}[/itex] is not sufficiently accurate for the squeeze theorem to assist, since [itex]\lim_{n \to \infty} \frac{n}{n-1} = 1[/itex].

But how, for [itex]n > 5[/itex], could you modify the above argument so that the squeeze theorem will assist?
 

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