It can be easily proven, $\forall p > 1$, that
$$\int_1^{\infty}\frac{dx}{x^p} = \frac{1}{p-1}.$$
By the integral test of convergence for infinite series, this implies both that $\sum\limits_{n=1}^{\infty} \frac{1}{n^p}$ converges by the same condition for $p$, and that
$$\sum_{n=1}^{\infty} \frac{1}{n^p} > \frac{1}{p-1}.$$
This establishes a strict lower bound for all p-series with $p > 1$.
With this in mind, does there exist an upper bound for the p-series, such that
$$\frac{1}{p-1} < \sum_{n=1}^{\infty} \frac{1}{n^p} < f(p)$$
where $f(p)$ is some positive function of $p$?
Thanks to @coffeemath for suggesting the visual proof of this.
So, the visual intuition for the integral test can be seen in the following image:
The area shown under the blue curve, $f(x)= \frac1{x^p}$, is $\frac1{p-1}$. Each rectangle has area $\frac1{n^p}$, where $n \in \mathbb{Z}^+$, thus the sum of the red rectangles represents $\sum\limits_{n=1}^\infty \frac1{n^p}$.
Since each rectangle is larger than the area underneath the curve bounded between it, this must mean that $\sum\limits_{n=1}^\infty \frac1{n^p} > \int\limits_1^\infty \frac{\text{d}x}{x^p} = \frac1{p-1}$.
Now, if ignore the first rectangle with area $1$, and shift each rectangle to the left by one unit, we get the following image:
Here, the sum of the rectangles represents $\frac1{2^p} + \frac1{3^p} + \frac1{4^p} + … = \left( \sum\limits_{n=1}^\infty \frac1{n^p} \right) - 1$, and they are clearly bounded below the same integral. Thus,
$$\left( \sum\limits_{n=1}^\infty \frac1{n^p} \right) - 1 < \frac1{p-1} \\ \implies \sum\limits_{n=1}^\infty \frac1{n^p} < 1 + \frac1{p-1} = \frac{p}{p-1} $$
Therefore, $\sum\limits_{n=1}^\infty \frac1{n^p}$ is bounded below $\frac{p}{p-1} \, \forall p > 1$.