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limn->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.

6 years ago


Answers : (1)


Dear Shraddhey

this type of question is solved by hepl of intigration

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

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

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

limn->∞ 1/n1/(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

=ob-a  dx/(a+x) 


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6 years ago

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