If $\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}.((na+1)+(na+2)+...+(na+n))}=\frac{1}{60}$, Find the value of a

\begin{align} \frac{1}{60}&=\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}.(n^2a+\frac{n(n+1)}{2})}\\ &=\lim_{n\to\infty}\frac{n^{a+1}\frac 1n\sum_{j=1}^n\Big(\frac jn\Big)^a}{(n+1)^{a-1}.(n^2a+\frac{n(n+1)}{2})}\\ &=\lim_{n\to\infty}\frac{n^{a+1}}{(n+1)^{a-1}.(n^2a+\frac{n(n+1)}{2})}\times \lim_{n\to\infty}\frac 1n\sum_{j=1}^n\Big(\frac jn\Big)^a\\ &=\frac{1}{a+\frac12}\int_0^1x^adx\\ &=\frac{1}{a+\frac12}\frac{1}{a+1} \end{align} where the integral requires $a>-1$. Therefore $a=7$ is the only acceptable solution.

Your first method is flawed. The limit is not zero.