limn→∞((n+1)(n+2).....3nn2n)1n is equal to:

Let the limit be equal to value A and apply log on both sides.logA=limn→∞log((n+1)(n+2).3nn2n)1n=limn→∞1nlog((n+1)(n+2).3nn2n)=limn→∞1nr=2n∑r=0log(n+rn)Taking1nasdxandrnasxLeta,bare the limits of the integration,⇒whenr=0,a=limn→∞1n=0; andr=2n,b=limn→∞2nn=2logA=limn→∞1nr=2n∑r=1log(1+rn)=∫20log(1+x)dx=[(1+x)log(1+x)−(1+x)]20[∵∫logx=x(logx𕒵)]logA=3log3𕒶A=e3log3𕒶=e3log3.e𕒶=e3log3e2=eloge27e2=27e2