Saturday, 3 September 2016

calculus - What is wrong with this fake proof that limlimitsnrightarrowinftysqrt[n]n!=1?



limnnn!=limnn1n2nn=111=1


I already know that this is incorrect but I am wondering why. It probably has something to do with the fact that multiplication in n! is done infinite number of times.


Answer



Start by figuring out a simpler example:
1=limnnn=limn1+1++1n=limn1n+1n++1n=0+0++0=0



Indeed, you cannot exchange sum (or product) and limit if the amount of terms in the sum or product depend on the limiting variable.


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