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

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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.**