Question

Find out how to answer this and the answer to this
$\frac{1}{2}$
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Answer 1

Answer:

My attempt :

limn→∞((n+1)(n+2)…(3n)n2n)1n

=limn→∞((n+1)(n+2)…(n+2n)n2n)1n

=limn→∞(n2n{(1+1/n)(1+2/n)…(1+2n/n)}n2n)1n

=limn→∞(n2n{(1+1/n)(1+2/n)…(1+2)}n2n)1n

=limn→∞(n2n{(1+1/n)(1+2/n)…(3)}n2n)1n

=limn→∞({(1+1/n)(1+2/n)…(3)})1n

I'm stuck here.