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.
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.
)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.
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.
20.25
To evaluate the limit, you can start by rewriting the given summation formula in terms of sigma notation:
∑((3i/n)^(3)*(3/n)), where i goes from 1 to n.
Next, we can simplify the expression inside the summation:
((3i/n)^(3))*(3/n) = (27i^3)/(n^4)
Now, we can rewrite the summation using the simplified expression:
∑((27i^3)/(n^4)), where i goes from 1 to n.
To evaluate the limit as n approaches infinity, we can use the fact that the sum of a series can be approximated by an integral when n is very large.
So, let's rewrite the summation as an integral:
∫((27x^3)/(n^4)) dx, where x goes from 1 to n.
Now, we can integrate the expression:
(27/n^4) * ∫(x^3) dx, where x goes from 1 to n.
Integrating x^3, we get:
(27/n^4) * [(x^4)/4], where x goes from 1 to n.
Now, let's evaluate the integral over the given interval:
(27/n^4) * [(n^4)/4 - (1^4)/4]
Simplifying further:
(27/4) * (1 - (1/n^4))
As n approaches infinity, the term (1/n^4) goes to zero. Therefore, the limit of the expression is:
(27/4) * (1 - 0) = 27/4
So, the value of the limit as n approaches infinity is 27/4.