Find limits of sequences

Solution for (b):

$$\lim_n\frac{1^p+3^p+\cdots +(2n-1)^p}{n^{p+1}}=\lim_n \frac{1^p+2^p+\cdots +(2n)^p}{n^{p+1}}-\lim_n\frac{2^p+4^p+\cdots+(2n)^p}{n^{p+1}}$$

$$=2^{p+1}\lim_n\frac{1^p+2^p+\cdots +n^p}{n^{p+1}}-2^p\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}$$

$$=(2^{p+1}-2^p).\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=2^p.\lim_n\frac{1}{n}\sum_{r=1}^n \left(\frac{r}{n}\right)^p$$

$$=2^p.\int_0^1x^p\,dx=\frac{2^p}{p+1}$$

We know that the sum of the $p^{th}$ powers of the integers is a polynomial of degree $p+1$ in $n$, let $S(n):=\alpha n^{p+1}+\beta n^p+R(n)$.

Then

$$n^p=S(n)-S(n-1)=\alpha n^{p+1}-\alpha (n-1)^{p+1}+\beta n^p-\beta (n-1)^p+R(n)-R(n-1)\\ =\alpha (p+1)n^p-\alpha \frac{(p+1)p}2n^{p-1}+\beta pn^{p-1}+Q(n)$$ where $Q$ is of degree at most $p-2$.

By identification,

$$\alpha =\frac1{p+1},\\\beta =\frac12$$ (which are the first two Faulhaber coefficients.)

This justifies the $\dfrac12$ in a).

For b), consider $$S(2n)-2^pS(n)=\frac{2^{p+1}n^{p+1}-2^pn^{p+1}}{p+1}+Q'(n)=\frac{2^pn^{p+1}}{p+1}+Q'(n)$$ to get the sum of the powers of the odd integers.

As you mentioned with Stolz–Cesàro theorem we will get

$$ b)\lim _{ n\to \infty } \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } =\lim _{ n\to \infty } \frac { 1^{ p }+3^{ p }+...+(2n+1)^{ p }-\left[ 1^{ p }+3^{ p }+...+(2n-1)^{ p } \right] }{ \left( n+1 \right) ^{ p+1 }-{ n }^{ p+1 } } =\\ =\lim _{ n\to \infty } \frac { { \left( 2n+1 \right) }^{ p } }{ { \left( p+1 \right) n }^{ p }+...+1 } =\lim _{ n\to \infty } \frac { { \left( 2n \right) }^{ p }+...+1 }{ { \left( p+1 \right) n }^{ p }+...+1 } =\frac { { 2 }^{ p } }{ p+1 } $$