Find this limit as x approaches infinity?

#lim_{xrarr \infty} (""^(2n)C_(n))^(1/n)#

1 Answer
Steve M
May 7, 2017

# lim_{n rarr oo} (""^(2n)C_(n))^(1/n) = 4#

Explanation:

I presume the question is meant to read #n rarr oo#, ie

# L = lim_{n rarr oo} (""^(2n)C_(n))^(1/n) #

Using the Combination Definition:

# ""^(n)C_r = (n!)/((n-r)!r!) #

We can write:

# L = lim_{n rarr oo} (((2n)!)/((2n-n)!n!))^(1/n) #
# \ \ = lim_{n rarr oo} (((2n)!)/(n!n!))^(1/n) #
# \ \ = lim_{n rarr oo} (((2n)!)/((n!)^2))^(1/n) #

For large #n# we can approximate #n!# using Stirling's approximation formula:

# n! ~~ sqrt(2 pi n) \ (n/e)^n #

Which gives us:

# L ~~ lim_{n rarr oo} ((sqrt(2 pi 2n) \ ((2n)/e)^(2n))/((sqrt(2 pi n) \ (n/e)^n)^2))^(1/n) #

# \ \ ~~ lim_{n rarr oo} ((sqrt(4 pi n) \ ((2n)/e)^(2n))/(2 pi n \ (n/e)^(2n)))^(1/n) #

# \ \ ~~ lim_{n rarr oo} ( (2sqrt(pi n)) /(2 pi n ) * (((2n)/e) / (n/e)) ^(2n))^(1/n) #

# \ \ ~~ lim_{n rarr oo} ( (2sqrt(pi n)) /(2 pi n ) * (2) ^(2n))^(1/n) #

# \ \ ~~ lim_{n rarr oo} ( 1/sqrt(pi n) )^(1/n) * lim_{n rarr oo}(4^n)^(1/n) #

# \ \ ~~ L_1 \ L_2 # , say

If we look at the first limit:

# L_1 = lim_{n rarr oo} ( 1/sqrt(pi n) )^(1/n) #
# \ \ \ \ = lim_{n rarr oo} ( (pi n)^(-1/2) )^(1/n) #
# \ \ \ \ = lim_{n rarr oo} (pi n)^(-1/(2n)) #
# \ \ \ \ = 1 #

And the second limit:

# L_2 = lim_{n rarr oo}(4^n)^(1/n) #
# \ \ \ \ = lim_{n rarr oo}4 = 4#

Combining these results we get:

# L ~~ 4 #

The approximation from Stirling's approximation formula becomes increasingly access as #n# increase, and as #n rarr oo#, we have:

# L = lim_{n rarr oo} (""^(2n)C_(n))^(1/n) = 4#

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