The exact question is- Find the real value(s) of $a (a \ne -1)$ for which the limit $$ \lim_{ n \to \infty} \frac{ 1^a+2^a\cdots+n^a}{(n+1)^{a-1}[ (na+1)+(na+2) \cdots+(na+n)]} = \frac{1}{60}$$
I simplified it a bit to get that limit. However, I got the answer from Wolfram Alpha $\frac{1}{a+1}$ without a solution and an assumption that $|a|<1$ however the answer that I got on further solving (and given) is $a=7$ and $a=-\frac{17}{2}$
How is it solved?
The expression is equal to
$$\frac{1}{n}\left( \left(\frac{1}{n}\right)^a +\cdots + \left(\frac{n}{n}\right)^a \right).$$
This is a Riemann sum, so your limit is the integral of an easy function.
Beside the Riemann sum, consider that $$\sum_{i=1}^n i^a=H_n^{(-a)}$$ which is a generalized harmonic number.
Using asymptotics $$H_n^{(-a)}=n^a \left(\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-a)$$ making for the question in title $$\frac{\sum_{i=1}^n i^a }{n^{a+1}}\sim \frac 1{a+1}+\frac a {2n}$$ So, the limit and also how it is approached.
For the question in text, $$\sum_{i=1}^n (na+i)=\frac{1}{2} \left((2 a+1) n^2+n\right)$$ and the same approach would lead to $$\sim \frac {n^a} {(n+1)^a} \frac{2(n+1)}{(1+a)(1+2a)n}\to \frac 2{(1+a)(1+2a) }$$
Useful link related to the first equation;
Evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}}$
Limit of the sequence $\frac{1^k+2^k+...+n^k}{n^{k+1}}$
$\lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=?$
Prove that limit of $\frac{1^p + 2^p + \ldots + n^p}{n^{p+1}}$ is equal to $\frac{1}{p+1}$
$$1^a+2^a+...+n^a\sim\frac{n^{a+1}}{a+1}$$ so we could write this limit :
$$ \lim_{ n \to \infty} \frac{ 1^a+2^a\cdots+n^a}{(n+1)^{a-1}[ (na+1)+(na+2) \cdots+(na+n)]} =\lim_{ n \to \infty}\frac{\frac{n^{a+1}}{a+1}}{{(n+1)^{a-1}[ n^2a+\frac{n(n+1)}{2}]}} = \lim_{ n \to \infty}\frac{2n^{a+1}}{{(a+1)n(n)^{a-1}n[2a+1]}} = \frac{2}{(a+1)(2a+1)}=\frac{1}{60}$$
So $(a+1)(2a+1)=120$ and $a=7,\frac{-17}{2}$
