For $a \in R-\{-1\}$, if $\lim\limits_{n \rightarrow \infty} \frac{\left(1^a+2^a+3^a+...+n^a\right)}{(n+1)^{a-1}\{(n a+1)+(n a+2)+...+(n a+n)\}}=\frac{1}{60}$, then $a=$

$7,-\frac{17}{2}$

We have,

$\lim\limits_{n \rightarrow \infty} \frac{\left(1^a+2^a+3^a+...+n^a\right)}{(n+1)^{a-1}\{(n a+1)+(n a+2)+(n a+3) ...+(n a+n)\}}=\frac{1}{60}$

$\Rightarrow \lim\limits_{n \rightarrow \infty} \frac{\left(1^a+2^a+3^a+...+n^a\right)}{(n+1)^{a-1}\left\{n^2 a+\frac{n(n+1)}{2}\right)}=\frac{1}{60}$

$\Rightarrow \lim\limits_{n \rightarrow \infty} \frac{2\left(\sum\limits_{r=1}^n r^a\right)}{n^{a+1}\left(1+\frac{1}{n}\right)^{a-1}\left(2 a+1+\frac{1}{n}\right)}=\frac{1}{60}$

$\Rightarrow 2\left\{\lim\limits_{n \rightarrow \infty} \frac{\frac{1}{n} \sum\limits_{r=1}^n\left(\frac{r}{n}\right)^a}{\left(1+\frac{1}{n}\right)^{a-1}\left(2 a+1+\frac{1}{n}\right)}\right\}=\frac{1}{60}$

$\Rightarrow 2\left\{\lim\limits_{n \rightarrow \infty} \frac{\frac{1}{n} \sum\limits_{r=1}^n\left(\frac{r}{n}\right)^a}{(2 a+1)}\right\}=\frac{1}{60}$

$\Rightarrow \frac{2}{2 a+1} \lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n} \sum\limits_{r=1}^n\left(\frac{r}{n}\right)^a\right\}=\frac{1}{60}$

$\Rightarrow \frac{2}{2 a+1}\left(\int\limits_0^1 x^a d x\right)=\frac{1}{60}$

$\Rightarrow \frac{2}{2 a+1}\left[\frac{x^{a+1}}{a+1}\right]_0^1$

$\Rightarrow \frac{2}{2 a+1}\left(\frac{1}{a+1}\right)=\frac{1}{60}$

$\Rightarrow 2 a^2+3 a+1=120$

$\Rightarrow 2 a^2+3 a-119=0$

$\Rightarrow 2 a^2+17 a-14 a-119=0$

$\Rightarrow (2 a+17)(a-7 b)=0$

$\Rightarrow a=7,-\frac{17}{2}$