Thursday, June 13, 2019

combinatorics - Evaluating limlimitsntoinftysumlimitsnk=0:(2n)kbinomnk



limnnk=0(nk)(2n)k
I've got to the form:
limn2n(2n1)(2n)n+11=limn2n+1n2n2n+1nn+11



And it should be e1/2 but I always get 0. I know it's just easy but I don't get it.



Answer



First use the binomial theorem:



nk=0(nk)(12n)k=(1+12n)n.



Now



limn(1+12n)n=limn((1+12n)2n)1/2=(limn(1+12n)2n)1/2,



and you should know what the last limit there is.



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