Limit of a sequence having positive integers raised to an exponent $a$ If $$lim_{n\rightarrow \infty} \frac{1^a + 2^a +3^a +\cdots +n^a}{(n+1)^{a-1}\left( (na+1)+(na+2) + \cdots +(na+n)\right)}= \frac{1}{60}$$
Find the value of $a$.
Obviously, the denominator turns out to be $$(n+1)^{a-1}(n^2a+\frac{n(n+1)}{2})$$
But I am not able to simplify the numerator.Can the numerator be simplified at all? What is the right way to solve this limit? 

This is not an answer but it is too long for a comment.
Interested by the post and impressed by the solution proposed by Daniel Fischer, I have been wondering if, for  finite values of $n$ and for any value of $a$ (integer, rational or not), we could obtain an approximation of
$$S_n=\frac{\sum_{i=1}^n i^a}{(n+1)^{a-1} \sum_{i=1}^n(na+i)}$$ Rewriting, as  Daniel Fischer did,
$$S_n=\frac{\sum_{i=1}^n i^a}{n^{a+1}} \cdot \frac{n^{a-1}}{(n+1)^{a-1}} \cdot \frac{n^2}{\sum_{i=1}^n(na+i)}=\frac{\sum_{i=1}^n i^a}{n^{a+1}} \cdot \left(\frac{n}{n+1}\right)^{a-1}\cdot\frac{2 n}{(2 a+1) n+1}$$ what I observed is that $$\frac{\sum_{i=1}^n i^a}{n^{a+1}}=\frac{1}{a+1}+\frac{1}{2 n}+\frac{a}{12
   n^2}+O\left(\frac{1}{n^3}\right)$$ The second and third terms were developed as Taylor series giving 
$$\left(\frac{n}{n+1}\right)^{a-1}=1+\frac{1-a}{n}+\frac{a(a-1)}{2 n^2}+\frac{a(1-a^2)}{6
   n^3}+O\left(\frac{1}{n^4}\right)$$
$$\frac{2 n}{(2 a+1) n+1}=\frac{2}{2 a+1}-\frac{2}{(2 a+1)^2 n}+\frac{2}{(2 a+1)^3 n^2}-\frac{2}{(2 a+1)^4
   n^3}+O\left(\frac{1}{n^4}\right)$$ Multiplying the three expansions leads to $$S_n=\frac{1}{(a+1) (2 a+1)}+\frac{-2 a^2+5 a+1}{(a+1) (2 a+1)^2 }\frac 1n+\frac{a(4 a^3-16 a^2+17 a-11) }{6 (a+1) (2 a+1)^3}\frac 1 {n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
For illustration purposes, using $n=a=5$ leads to an exact value of $\frac{295}{12096}\approx 0.0243882$ while the above approximation gives $\frac{65}{2662}\approx 0.0244177$.
