<s>[INST] Evaluate this question based on series and limits. 
For $a \in \mathbb R,a≠-1$
$$\lim_{n\to\infty}\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}=\frac{1}{60}$$
  Then find the values of $a $.

I tried to solve this problem using approximation but I got the the value of $a$ as $\dfrac{-3}{2}$. The source of this question is JEE advanced $2013$ of India.
 [/INST] For $a>-1$ and $a\neq -\frac 12$ we have
\begin{align}
\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}
&\sim\frac {n^{a+1}/(a+1)}{n^{a-1}(a+1/2)n^2}\\
\to\frac 2{(a+1)(2a+1)}
\end{align}
from which $a=7$.
For $a<-1$ the numerator converges to  positive value, hence the limit is $-\infty $.
Finally, for $a=-\frac 12$ we have
\begin{align}
\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}
&\sim\frac {2\sqrt n}{n^{-3/2}n/2}\\
\sim 4n\to +\infty
\end{align}
The answer $a=7$ is confirmed:

</s>