Proving the limit of a sequence

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  • #1
silvermane
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Homework Statement


Prove that the limit of 2n/n! = 0

The Attempt at a Solution


I've already successfully proven that 2n < n! for every n>4, so for every value of xn greater than 4, we know that 0<xn<1.

I understand how the sequence is behaving, but I'm having trouble putting it into words with an analytical mindset. Any hints or tips would be greatly appreciated!!
:blushing:
 

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  • #2
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Homework Statement


Prove that the limit of 2n/n! = 0

The Attempt at a Solution


I've already successfully proven that 2n < n! for every n>4, so for every value of xn greater than 4, we know that 0<xn<1.

I understand how the sequence is behaving, but I'm having trouble putting it into words with an analytical mindset. Any hints or tips would be greatly appreciated!!
:blushing:
Fix an ∂ > 0, and then you're seeking an N such that

|2n/n! - 0| = 2n/n! < ∂

whenever n ≥ N.

Alternatively,

n!/2n > 1/∂.

Because n!/2n > n whenever n ≥ 6,

It is sufficient to show

n > 1/∂

whenever n > N.

Let N be the smallest integer greater than 1/∂ (Archimedean Property) and then N, N+1, N+2, ... also be greater than 1/∂.

Thus there does exist an N such that

2n/n! < ∂

whenever n ≥ N.
 
  • #3
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If you use ratio test for the infinite sum of 2n/n!, it converges and so 2n/n! tends to 0. That's probably the easiest way to do it, although it gets into series.
 
  • #4
jbunniii
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Homework Statement


Prove that the limit of 2n/n! = 0

The Attempt at a Solution


I've already successfully proven that 2n < n! for every n>4
Good, you can use that fact.

Note that for [itex]n > 0[/itex],

[tex]\frac{2^n}{n!} = \frac{2^{n-1}}{(n-1)!} \cdot \frac{2}{n}[/tex]

You showed that

[tex]\frac{2^{n-1}}{(n-1)!} < 1[/tex]

for large enough [itex]n[/itex], so use that fact to bound this sequence by one whose limit you know.
 
  • #5
silvermane
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If you use ratio test for the infinite sum of 2n/n!, it converges and so 2n/n! tends to 0. That's probably the easiest way to do it, although it gets into series.
We're not permitted to use this test. It's an analysis course where we must use the definition of the limit at this point and time :(
 
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  • #6
silvermane
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Okay, this is what I ended up doing: (thanks to everyone of course)

Let N be the smallest natural number such that N>2. then, for n greater/equal N,

2n/n! is less/equal to [tex]\frac{2}{1}[/tex]*[tex]\frac{2}{2}[/tex]*[tex]\frac{2}{N-1}[/tex]*[tex]\frac{2}{N}[/tex]*...*[tex]\frac{2}{N}[/tex]

= [tex]\frac{2^(N-1)}{(N-1)!}[/tex]*[tex]\frac{2}{N}[/tex]^(n-N+1)

Since 2/N <1, (2/N)n goes to 0 as n goes to infinity, and it follows from the squeeze theorem.
 

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