lim_n->infinty [1/na + 1/(na+1) + 1/(na+2) + ...... + 1/nb]

Plz show the method. Ans is -> log (b/a).

This question is from arihant, last workbook exercise of Limits.

Dear Shraddhey

this type of question is solved by hepl of intigration

lim_n->∞ [1/na + 1/(na+1) + 1/(na+2) + ...... + 1/(na+nb-na)]

lim_n->∞ ∑1/(na+r)               r  vary from 0 to (b-a)

lim_n->∞ 1/n ∑n/(na+r)               r  vary from 0 to (b-a)

lim_n->∞ 1/n ∑1/(a+r/n)               r  vary from 0 to (b-a)

now if any given series is of this form then replace ∑  by ∫   ,and  r/n   by  x , and  1/n by dx

=_o∫^b-a  dx/(a+x) 

=lnb/a

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin