Answer:
Option D
Explanation:
We have
yn=1n((n+1)(n+2)...(n+n))1n and
limn→∞yn=L
⇒ L= limn→∞1n[(n+1)(n+2)(n+3)....(n+n)]1n
⇒ L= limn→∞1n[(1+1n)(1+2n)(1+3n)....(1+nn)]1n
⇒ logL=limn→∞1n[log(1+1n)+log(1+2n)+log(1+3n)....log(1+nn)]
⇒ logL=limn→∞1n∑nrlog(1+rn)
⇒ logL=∫101×log(1+x)dx
⇒ logL=(x.log(1+x)10−∫10[ddx(log(1+x)∫dx]dx
[ by using integration by parts]
⇒ logL=[xlog(1+x)10−∫10x1+xdx
⇒ logL=log2−∫10(x+1x+1−1x+1)dx
⇒ logL=log2−[x]10+[log(x+1)]10
⇒ logL=log2−1+log2−0
⇒ logL=log4−loge=log4e
⇒ L=4e
⇒ [L]=[4e]=1