# Calculus (integrals)

posted by on .

Use the following formula for the sum of the cubes of the first integers to evaluate the limit in part (a).
1**3+2**3+...+n**3=((n(n+1))/2)**2

(a)lim n approaches infinity and the sum of n (top) and i=1 (bottom) with (3i/n)^3*(3/n)

I don't know how to solve this, can you help me out? Thank you.

• Do not understand - ,

Please use ^ to denote exponent

I can not figure out what your exponent really is in the summation and what you want the limit of.

• Calculus (integrals) - ,

)lim n approaches infinity and the sum of n (top) and i=1 (bottom) with (3i/n)^3*(3/n)

sum from i = 1 to i =n of what?

(3i/n)^3*(3/n)what is this?

sum from i = 1 to oo of
(3i/n)^ [3*(3/n) ]
is
sum from i = 1 to oo of
(3i/n)^ [9/n]
is that what you mean? I think not but I have no idea.

• series - ,

If I understand correctly, the problem is to find the limit of the sum for i=1 to ∞ for the expression:

(3i/n)³(3/n)
=3³ i³ (3/n^4)
=(3/n)^4 (i)³
=(3/n)^4 ( i(i+1)/2 )²
=3^4/2^sup2; ( (i^4+2i³+i²)/n^4 )
=3^4/2^sup2; (i^4/n^4 + 2i³/n^4 + i²/n^4;)

If we take the limit as i->∞ we end up with the sum of
3^4/2^sup2;

Check my thinking.

• Calculus (integrals) - ,

20.25