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.
On
Let $$S=\lim_{n\to\infty}\frac{\sum_{r=0}^nr^a}{(n+1)^{a-1}\sum_{r=0}^n ( na + r)} $$
Observe that the sum can be converted to a reimann integral as,
$$S = \lim_{n\to\infty} \frac{2}{n}\sum_{r=1}^n \frac{(\frac{r}{n})^a}{(2a+\frac{1}{n} + 1)(1+\frac{1}{n})^{a-1}}$$
$$ S = 2\int_0^1 \frac{x^a}{2a+1}\mathrm dx$$ $$ S = \frac{2}{(2a+1)(a+1)} = \frac{1}{60} \implies a = 7$$
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: