IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

agnit_thebest

(limit n tends to infinity) [1/(2.3)+1/(3.5)+1/(5.7)+1/(7.9)+......to n terms]

the answer is 0.5.

iitkgp_bipin

I think the expression is :

1/1.3 + 1/3.5 + 1/5.7 + ......+ 1/(2n-1).(2n+1)

= (1/2)(1/1 - 1/3) + (1/2)(1/3 - 1/5) + (1/2)(1/5 - 1/7) + ......+(1/2)(1/2n-1 - 1/2n+1)

If we tale (1/2) common and add all the terms they cancel out except 1 and -1/2n+1.

Sum = (1/2)(1 - 1/2n+1)

When n tends to infinity sum tends to 1/2.

aditya_arora04

I to agree with sir

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