Prove rigorously: lim (3^n)/(n) = 0

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In summary, the problem is to prove that the limit of (3n)/(n!) as n approaches infinity is equal to 0. The definition states that a real number sequence a(n) approaches a limit a if for all epsilon greater than 0, there exists an integer N such that n is greater than or equal to N implies that the absolute value of a(n) minus a is less than epsilon. The discussion shows that the limit can be proven to be 0 by finding an N such that the terms in the sequence decrease by a factor of 1/2 each time, ultimately becoming smaller than epsilon.
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kingwinner
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Homework Statement


PROVE rigorously from the definition that
lim (3n)/(n!) = 0.
n->∞

Homework Equations


N/A

The Attempt at a Solution


By definition, a real number sequence
a(n)->a iff
for all ε>0, there exists an integer N such that n≥N => |a(n) - a|< ε.

|(3n)/(n!)|<...< ε
Now how can I find N? The usual approach to find N would be to set |a(n) -L|< ε and solve the inequality for n. But here in |(3n)/(n!)|<...< ε, I don't think we can solve for n. (because n is appearing everywhere. n(n-1)(n-2)...2x1, and n also appears on the numerator)

Any help is greatly appreciated!



[note: also under discussion in Math Links forum]
 
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Oh, come on. 3^6/6!=81/80. The next term multiplies that by 3/7. That factor is less than 1/2. The next term multiplies that by 3/8, that's also less than 1/2. The next term multiplies that by 3/9, also less than 1/2. Can you find an N such that (81/80)*(1/2)^N is less than epsilon? I think you probably can. Call off the debate.
 

FAQ: Prove rigorously: lim (3^n)/(n) = 0

What is the definition of a limit in calculus?

The limit of a function is the value that the function approaches as the input approaches a certain value or approaches infinity. It represents the behavior of the function near a specific point.

How can we prove that the limit of a function exists?

To prove that the limit of a function exists, we can use the epsilon-delta definition which states that for any small positive number epsilon, there exists a small positive number delta such that if the distance between the input and the specific point is less than delta, then the distance between the output and the limit is less than epsilon.

How do we prove that the limit of a sequence is equal to 0?

To prove that the limit of a sequence is equal to 0, we can use the definition of a limit for sequences which states that for any small positive number epsilon, there exists a positive integer N such that if the index of the term is greater than N, then the term itself is less than epsilon.

What is the proof for the limit of (3^n)/(n) as n approaches infinity equal to 0?

To prove that the limit of (3^n)/(n) as n approaches infinity is equal to 0, we can use the definition of a limit for sequences. Let epsilon be a small positive number. Then, we can choose N = 1/epsilon as our positive integer. Now, for any index n greater than N, we have (3^n)/(n) < (3^n)/(1/epsilon) = epsilon. Therefore, the limit of (3^n)/(n) as n approaches infinity is equal to 0.

Can we use other methods to prove that the limit of (3^n)/(n) as n approaches infinity is equal to 0?

Yes, we can also use the squeeze theorem or the ratio test to prove that the limit of (3^n)/(n) as n approaches infinity is equal to 0. Additionally, we can use the definition of a limit for functions and show that the limit of (3^n)/(n) as n approaches infinity is equal to 0 by showing that the distance between the function and the limit can be made arbitrarily small for a given input n.

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