The argument below works for any real $a > -1$. We are given that $s_n(a) = \sum_{k=1}^{n} k^a$ Let $a_n = 1$ and $A(t) = \displaystyle \sum_{k \leq t} a_n = \left \lfloor t \right \rfloor$. Hence, $s_n(a) = \int_{1^-}^{n^+} t^a dA(t)$ The integral is to be interpreted as the Riemann Stieltjes integral. Now integrating by parts, we get that $s_n(a) = \left. t^a A(t) \right \rvert_{1^-}^{n^+} - \int_{1^-}^{n^+} A(t) a t^{a-1} dt = n^a \times n - a \int_{1^-}^{n^+} \left \lfloor t \right \rfloor t^{a-1} dt\\ = n^{a+1} - a \int_{1^-}^{n^+} (t -\left \{ t \right \}) t^{a-1} dt = n^{a+1} - a \int_{1^-}^{n^+} t^a dt + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ = n^{a+1} - a \left. \dfrac{t^{a+1}}{a+1} \right \rvert_{1^-}^{n^+} + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ =n^{a+1} - a \dfrac{n^{a+1}-1}{a+1} + a \int_{1^-}^{n^+}\left \{ t \right \} t^{a-1} dt\\ = \dfrac{n^{a+1}}{a+1} + \dfrac{a}{a+1} + \mathcal{O} \left( a \times 1 \times \dfrac{n^a}{a}\right)\\ = \dfrac{n^{a+1}}{a+1} + \mathcal{O} \left( n^a \right)$ Hence, we get that $\lim_{n \rightarrow \infty} \dfrac{s_n(a)}{n^{a+1}/(a+1)} = 1$ Hence, now $\dfrac{s_{n}(a+1)}{n s_n(a)} = \dfrac{\dfrac{s_n(a+1)}{n^{a+2}/(a+2)}}{\dfrac{s_n(a)}{n^{a+1}/(a+1)}} \times \dfrac{a+1}{a+2}$ Hence, we get that $\lim_{n \rightarrow \infty} \dfrac{s_{n}(a+1)}{n s_n(a)} = \dfrac{\displaystyle \lim_{n \rightarrow \infty} \dfrac{s_n(a+1)}{n^{a+2}/(a+2)}}{\displaystyle \lim_{n \rightarrow \infty} \dfrac{s_n(a)}{n^{a+1}/(a+1)}} \times \dfrac{a+1}{a+2} = \dfrac11 \times \dfrac{a+1}{a+2} = \dfrac{a+1}{a+2}$
Note that the argument needs to be slightly modified for $a = -1$ or $a = -2$. However, the two cases can be argued directly itself.
If $a=-1$, then we want $\lim_{n \rightarrow \infty} \dfrac{s_n(0)}{n s_n(-1)} = \lim_{n \rightarrow \infty} \dfrac{n}{n H_n} = 0$
If $a=-2$, then we want $\lim_{n \rightarrow \infty} \dfrac{s_n(-1)}{n s_n(-2)} = \dfrac{6}{\pi^2} \lim_{n \rightarrow \infty} \dfrac{H_n}{n} = 0$
In general, for $a <-2$, note that both $s_n(a+1)$ and $s_n(a)$ converge. Hence, the limit is $0$. For $a \in (-2,-1)$, $s_n(a)$ converges but $s_n(a+1)$ diverges slower than $n$. Hence, the limit is again $0$.
Hence to summarize $\lim_{n \rightarrow \infty} \dfrac{s_n(a+1)}{n s_n(a)} = \begin{cases} \dfrac{a+1}{a+2} & \text{ if }a>-1\\ 0 & \text{ if } a \leq -1 \end{cases}$