First, for $a>-1$: $$\lim_{n\to\infty}\frac{a+1}{n^{a+1}}\sum_{j=1}^nj^a = 1$$
Second, for $p>0$: $$\lim_{n\to\infty}\frac{e^a-1}{e^{a(n+1)}}\sum_{j=1}^ne^{aj} = 1$$
In particular, why do we need the restriction on constants?
Lastly, what is the normalizing factors $s_{1n}, s_{2n}, s_{3n}$ (as functions of $a,b,n$) for $S_{1n} = \sum_{j=1}^nj^ae^{bj}$, $S_{2n} = \sum_{j=1}^nj^ae^{j^b}$ and $S_{3n} = \sum_{j=1}^ne^{aj + j^b}$ and restrictions on $a,b$, so that $$\lim_{n\to\infty}s_{jn}S_{jn} = 1,\quad j=1,2,3$$
The only thing that I see is that $\frac{1}{n^{a+1}}\sum_{j=1}^nj^a \leq 1$, but this obviously does not give me a limit.
You need restrictions on the constants because the limits vary depending on where $a$ lies. For the first limit,
$$\lim_{n\to \infty} \frac{a + 1}{n^{\alpha + 1}}\sum_{j = 1}^n j^a = (a + 1)\lim_{n\to \infty} \frac{1}{n}\sum_{j = 1}^n \left(\frac{j}{n}\right)^a = (a + 1)\int_0^1 x^a\, dx = x^{a+1}\bigg|_{x = 0}^1 = 1.$$
For $a > 0$, $\lim_{n\to \infty} e^{-a(n+1)} = 0.$ So
$$\lim_{n\to \infty} \frac{e^a - 1}{e^{a(n+1)}}\sum_{j = 1}^n e^{aj} = \lim_{n\to \infty} \frac{e^a - 1}{e^{a(n+1)}}\frac{e^{a(n+1)} - e^a}{e^a - 1} = \lim_{n\to \infty} (1 - e^{-an}) = 1.$$