Solve limit of y=[1+3^4+5^4+...+(2n-1)^4]/n^5 if applying the Cesaro-Stoltz theorem.

The Stolz-Cesaro's rule states that is 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 = 1+3^4+5^4+...+(2n-1)^4

un+1 =
1+3^4+5^4+...+(2n+2-1)^4

un+1 =
1+3^4+5^4+...+(2n+1)^4

un+1 - un = 1+3^4+5^4+...+(2n+1)^4 -
1-3^4-5^4-...-(2n-1)^4

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

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

We'll put vn =
n^5

vn+1 = (n+1)^5

vn+1 - vn =
(n+1)^5 - n^5

We'll have to prove that the limit of the
ratio (2n+1)^4/[(n+1)^5 - n^5] exists.

We'll expand the
binomial from numerator:

(2n+1)^4 = 16n^4 +
...

We'll expand the binomial from
denominator:

(n+1)^5 = n^5 + 5n^5 +
...

We'll subtract
n^5:

[(n+1)^5 - n^5] = n^5 + 5n^4 + ... -
n^5

[(n+1)^5 - n^5] = 5n^4 +
...

We'll re-write the
limit:

lim (2n+1)^4/[(n+1)^5 - n^5] = lim (16n^4 +
...)/(5n^4 + ...)

We'll factorize by
n^4:

lim n^4(16 + ...)/n^4(5 + ...) =
16/5

Since the limit of the ratio
(2n+1)^4/[(n+1)^5 - n^5] = 16/5, then the limit of the ratio
[1+3^4+5^4+...+(2n-1)^4]/n^5 is 16/5, too.