evalute 1/n^100 lim as r tends to infinity find summation ( r goes from 1 to n ) ( r^99)

Akhil

Deepak Aggarwal

Ans is 1/100..........

Akhil

HOW TO WE GET THAT RESULT..

COULD YOU PLS PROVE THAT RESULT

Filokalk d' nix

another way is to use integration for limit of sum

S= lim (n->infinity) summation (1 to n){ (r/n)^99 * 1/n}

= integral(0 to 1) x^99 dx

=[(x^100)/100]..........( use limits 0&1

=1/100

vijay kharya

@ filokalk shouldn't the limits used be from 1 to n because the summation is from 1 to n.