Evaluate the limit of string [2^2+4^2+6^2+...+(2n-2)^2]/n^3 if the number of terms of string is infinite?

To evaluate this limit, we'll need the Cezaro-Stolz's
theorem. This theorem states that if the limit of the ratio (un+1 - un)/(vn+1 - vn)
exists, then the limit of the ratio un/vn exists  and it is equal to the previous
limit.

We'll put the sum from numerator as un =
2^2+4^2+6^2+...+(2n-2)^2

un+1 =2^2+4^2+6^2+...+(2n-2)^2 +
[2(n+1) - 2]^2

un+1 = 2^2+4^2+6^2+...+(2n-2)^2 +
(2n)^2

un+1 - un = 2^2+4^2+6^2+...+(2n-2)^2 + (2n)^2 - 2^2
- 4^2 - 6^2-...- (2n-2)^2

We'll eliminate like terms and
we'll get:

un+1 - un =
(2n)^2

We'll put vn = n^3

vn+1
= (n+1)^3

vn+1 - vn = (n+1)^3 -
n^3

We'll have to prove that the limit of the ratio
(2n)^2/[(n+1)^3 - n^3] exists.

We'll expand the cube from
denominator:

(n+1)^3 = n^3 + 3n^2 + 3n +
1

We'll subtract n^3:

[(n+1)^3
- n^3] = n^3 + 3n^2 + 3n + 1- n^3

[(n+1)^3 - n^3] = 3n^2 +
3n + 1

We'll re-write the
limit:

lim (2n)^2/[(n+1)^3 - n^3] = lim(2n)^2/(3n^2 + 3n +
1)

We'll factorize by n^2:

lim
2n^2/n^2(3 + 3/n + 1/n^2) = lim 2/(3 + 3/n + 1/n^2)

lim
2/(3 + 3/n + 1/n^2) = 2/3, if n approaches to
infinite

Since the limit of the ratio
(2n)^2/[(n+1)^3 - n^3] = 2/3, then the limit of the ratio [2^2+4^2+6^2+...+(2n-2)^2]/n^3
= 2/3.