Posted by **Sarah** on Friday, April 16, 2010 at 6:52pm.

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 -
**Damon**, Friday, April 16, 2010 at 7:10pm
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) -
**Damon**, Friday, April 16, 2010 at 7:15pm
)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 -
**MathMate**, Friday, April 16, 2010 at 7:46pm
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) -
**Naumair**, Saturday, April 17, 2010 at 3:38pm
20.25

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