Let's do some calculus! (18)

$\lim_{n\to \infty} \dfrac{\sum_0^n k^a}{n^{a+1}}=\int_0^1 x^a dx=\dfrac{1}{a+1}$ we can write the limit as $\lim_{n\to \infty} \dfrac{\sum_0^n k^a}{(n+1)^{a-1} (n^2a+n(n+1)/2)}=\lim_{n\to \infty} \dfrac{\sum_0^n k^a}{n(n+1)^{a-1} ((a+1/2)n+1/2)}=\dfrac{1}{(a+1/2)(a+1)}$ the last step was possible because we ignored all other terms in the denominator but the leading term, which is $(a+1/2)n^{a+1}$ , and the other part we already evaluated at the beginning of this solution. from this we get the equation $(a+1)(a+1/2)=60\to a=\boxed{7}, -8.5$

The denominator is readily evaluated using sum of first n natural numbers.

For evaluating Numerator divide by n^a in numerator and denominator.

The numerator is readily evaluated via Definite Integration.

The denominator is simply evaluated by taking n towards infinity.

Integrating and solving we get a quadratic in a

2a^2+3a-119 = 0

Which has root a = 7 and -17/2

But since a is positive we reject a = -17/2