Limit $\lim_{n \rightarrow \infty} \frac {1^{a+1}+2^{a+1}+\cdots+n^{a+1}}{n.(1^{a }+2^{a }+\cdots+n^{a })}$ The value of $$\lim_{n=\infty} \dfrac {1^{a+1}+2^{a+1}+\cdots+n^{a+1}}{n.(1^{a }+2^{a }+\cdots+n^{a })}  $$
Attempt: $S = \lim_{n \rightarrow \infty} \sum_{n=0} ^\infty \dfrac {k^{a+1}} {n.( 1^{a }+2^{a }+\cdots+ n^{a} )}$
I am trying to use the integral as a limit of sum but the denominator is preventing me from applying.
Could anyone please give me a direction to move on.
Thank you for your help.

The limit can be interpreted as the ratio of two Riemann sums. First of all we have that
$$I_k = \lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n \left(\frac{i}{n}\right)^k = \int_0^1 x^k dx = \frac{1}{k+1}$$
which gives $$S = \lim_{n\to \infty}\frac{\frac{1}{n}\sum_{i=1}^n \left(\frac{i}{n}\right)^{a+1}}{\frac{1}{n}\sum_{i=1}^n \left(\frac{i}{n}\right)^{a}} = \frac{I_{a+1}}{I_a} = \frac{a+1}{a+2}$$
where in the second step above I have used the fact that if $a_n$ and $b_n$ converges then $\lim_{n\to\infty} \frac{a_n}{b_n} = \frac{\lim_{n\to\infty} a_n}{\lim_{n\to\infty} b_n}$.
