<s>[INST] 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? 
 [/INST] 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 $\frac{65}{2662}\approx 0.0244177$.
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