Evaluate the limit $ \lim_\limits{n \to \infty} \frac{1^a+2^a\cdots+n^a}{n^{a+1}} $ 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?

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 $\frac{1^k+2^k+...+n^k}{n^{k+1}}$
$\lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=?$
Showing that $\lim_{n \rightarrow \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}=\frac{1}{k+1}$ using integral
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 $a=7,\frac{-17}{2}$
