lim n → ∞ 1 n + n n + 1 2 + n n + 2 2 + … + n 2 n - 1 2 is equal to

limn→∞1n+nn+12+nn+22+…+n2n-12 is equal to
  1. 12
  2. 14
  3. 13
  4. 1

Solution

limn→∞1n+nn+12+nn+22+…+n2n-12

=limn→∞∑r=0n-1nn+r2=limn→∞∑r=0n-1nn2+2nr+r2

=limn→∞1n∑r=0n-11r/n2+2r/n+1

=∫01dxx+12=-1x+101=12

Asked in: JEE Main 2021 (25 Feb Shift 2)

Practice more Definite Integration questions on Aicharya