limits problem

[rongmon]

The limiting value of [1.2 + 2.3 + 3.4 + .... n(n+1)]/n^3 as n tends to infinity is

a. 0
b. 1
c. 1/3
d. 1/2

thanks

[Deepak_Ka]

find the general term which in this case is tn=n(n+1).
use summation formula for n2 and n.

answer is 1/2.

[rongmon]

i have the answer as 1/3

[Noel]

[rongmon]

thanks a ton

[Ashima]

Yeah 1/3! You can do it simply by L' hopital rule.

[rongmon]

What term would you differentiate at the numerator?

[Ashima]

I am getting n^3+3n^2+2n/3n^3
= 3n^2+6n+2/ 9n^2= 6n+6/18n= 6/18= 1/3

Tell me if i am wrong.

[Gayyam]

How would you apply L'Hospital's rule? The rule requires that the function be differentiable. Here the function that you have taken is defined only for natural numbers.

Although you may get the same answer, the method is unsound without further justification.

[Dreyfus]

Instead of taking derivative which is isn't possible divide each term with n^3....nd when n→ infinity the whole expression reduces to 2/6....

[Ashima]

Here n is variable. n^3 , n^ 2 and n all are differentiable! Tell me why cant I use L' hospitals?

[Noel]

The domain of all these functions is the set of natural numbers..which is a discrete set..so the range set cant be continuous here..hence it cant be differentiable..

[Ashima]

Acha its the nth term. I was taking it as a variable I see ur point. tHANKS :)