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? 
 A: There is no closed formula for $1^a + 2^a + \dotsc + n^a$ with general $a$, but the asymptotic behaviour of that sum is easy enough, and the asymptotic behaviour is all that matters here. Show first that the sequence is unbounded for $a \leqslant -1$. For $a > -1$, write the term as
$$\frac{1^a + 2^a + \dotsc + n^a}{n^{a+1}} \cdot \frac{n^{a-1}}{(n+1)^{a-1}} \cdot \frac{n^2}{(na+1)+(na+2)+\dotsc +(na+n)}.$$
Each factor in that has an easy to find limit. That gives a simple equation to solve for $a$.
A: 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 $\frac{65}{2662}\approx 0.0244177$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\chi\pars{a} \equiv \lim_{n \to \infty}{\sum_{k = 1}^{n}k^{a} \over
\pars{n + 1}^{a - 1}\sum_{k = 1}^{n}\pars{na + k}} =
\lim_{n \to \infty}{\sum_{k = 1}^{n}k^{a} \over
\pars{n + 1}^{a - 1}\,n^{2}\bracks{a + 1/2 + 1/\pars{2n}}}
:\ {\large }}$


$\ds{\Large a < -1:}$
\begin{align}
&{\sum_{k = 1}^{n}k^{a} \over
\pars{n + 1}^{a - 1}\,n^{2}\bracks{a + 1/2 + 1/\pars{2n}}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{\zeta\pars{-a} \over a + 1/2}\,n^{-a - 1}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, -\infty
\\[5mm] & \implies
\bbx{\mbox{There}\ \mathbf{isn't}\ \mbox{any solution for}\ a < -1}
\end{align}

$\ds{\Large a = -1:}$
\begin{align}
&{\sum_{k = 1}^{n}k^{a} \over
\pars{n + 1}^{a - 1}\,n^{2}\bracks{a + 1/2 + 1/\pars{2n}}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
-2\,H_{n}\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, -\infty
\\[5mm] & \implies
\bbx{a = - 1\quad \mathbf{isn't}\quad \mbox{a solution}}
\end{align}

$\ds{\Large -1 < a < 0:}$
\begin{align}
\sum_{k = 1}^{n}k^{a} & = \sum_{k = 1}^{n}{1 \over k^{-a}} =
{n^{1  + a} \over 1 + a} + \zeta\pars{-a} -
a\int_{n}^{\infty}{\braces{x} \over x^{-a + 1}}\,\dd x
\end{align}


Note that
  $\ds{\int_{n}^{\infty}{\braces{x} \over x^{-a + 1}}\,\dd x < \int_{n}^{\infty}x^{a - 1}\,\dd x = -\,{1 \over an^{-a}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {\large 0} }$

\begin{align}
&{\sum_{k = 1}^{n}k^{a} \over
\pars{n + 1}^{a - 1}\,n^{2}\bracks{a + 1/2 + 1/\pars{2n}}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,
\left\{\begin{array}{lcl}
\ds{1 \over \pars{1 + a}\pars{a + 1/2}} & \mbox{if} &
\ds{a \not= -\,{1 \over 2}}
\\[2mm]
\ds{\infty} & \mbox{if} & \ds{a = -\,{1 \over 2}}
\end{array}\right.
\end{align}

$\ds{{1 \over \pars{1 + z}\pars{z + 1/2}} = {1 \over 60} \implies z = -\,{17 \over 2}\ \mbox{or}\ z = 7}$ but they don't belong to $\ds{\pars{-1,0}}$.


$\ds{\Large a = 0:}$ In this case, it's obvious that
$\ds{\chi\pars{0} = \infty}$.

$\ds{\Large a > 0:}$ The limit is trivially reduced to the evaluation of two Riemann sums. Namely,
$$
{\int_{0}^{1}x^{a}\,\dd x \over \int_{0}^{1}\pars{a + x}\,\dd x} =
{1 \over \pars{a + 1}\pars{a + 1/2}} = {1 \over 60}
$$
It leads to the solution
$\bbox[10px,#ffe,border:1px dotted navy]{\Large\ds{a = 7}}$ because another "possible" solution
$\ds{\pars{~z = \require{cancel}\cancel{-\,{17 \over 2}}~}}$ does not belong to $\ds{\pars{0,\infty}}$.
A: For $a>0$ and $j\geq 1$ we have  $$\int_{j-1}^jx^adx <j^a<\int_j^{j+1}x^adx.$$ Summing from $j=1$ to $j=n$ gives $$n^{a+1}/(a+1)=\int_0^nx^adx<(1^a+...+n^a)<\int_1^{n+1}x^adx=((n+1)^{a+1}-1)/(a+1)$$ which is a sufficiently good estimate for the numerator  to determine that  $a>0 \implies (a+1)(a+1/2)=60\implies a=7.$
