limn→∞n∑k=0(nk)(2n)k
I've got to the form:
limn→∞2n(2n−1)(2n)n+1−1=limn→∞2n+1n−2n2n+1nn+1−1
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:
n∑k=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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