Limits and Factorials!
n → ∞ lim n 1 i = 1 ∑ n ( i ! ) 1 / i 2 = ?
Give your answer to 3 decimal places.
Submit your answer as -0.222 if you think the limit diverge.
The answer is 1.
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2 solutions
First note that, n → ∞ lim ( n ! ) 1 / n 2 = 1 This result follows easily from Stirling's approximation . Finally, we invoke the Cesaro mean theorem to complete the proof.
Is there any other method other than this?
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It's much simpler if you apply Stolz–Cesàro theorem :
If a n and b n are two sequences of real numbers and b n is strictly monotone and divergent, and the limit L = n → ∞ lim b n + 1 − b n a n + 1 − a n exists, then n → ∞ lim b n a n also exists and it's equal to L .
In this case, a n = i = 1 ∑ n ( i ! ) 1 / n 2 and b n = n , then L = n → ∞ lim b n + 1 − b n a n + 1 − a n simplifies to n → ∞ lim ( n + 1 ) ! 1 / ( n + 1 ) 2 = n → ∞ lim n ! 1 / n 2 = 1 which can be evaluated using Stirling's formula to 1, and it agrees with Abhishek's result. Thus L = 1 and so is the answer in question.
What about sandwitch theorem ....
On left side each term more than 1 /n...
on right side each term more than 2r/n ....
Right hand side evaluated using limit of sums....
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Sorry right hand side each term... 2r/n^{2}
By Cauchy's First limit theorem we know that the arithmetic mean of a sequence converges to the same limit as that of the sequence provided the limit of the sequence exists .....for proof see the solution of this problem : - https://brilliant.org/problems/limit-of-a-sequence-07/ So we just need to find lim (n!)^1/n^2. Which we can do either by Stirling's approximation or with the help of Cauchy's 2nd limit theorem which is a corollary of the first theorem that states that (xn)^1/n converges to the same limit as lim xn+1 / xn (the limit of the ratio of the n+1 th term and the nth term) provided that the limit existis. Anyways with the help of anyone of the theorems above we get that the limit is 1. So the Arithmetic mean also converges to the same limit