wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate limn(1n2+1+1n2+2+1n2+3+...+nn2+n)

Open in App
Solution

limn(1n2+1+1n2+2+1n2+3+......+nn2+n)

Now dividing both numerator and denominator by n2
limn1n21+1n2+1n21+2n2+1n21+3n2+......+1n21+1n
=01+01+01........01
=0

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Mathematical Induction
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon