Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$

We can bound the partial sums by multiples of themselves:

$$\begin{eqnarray} S_{2k+1} &=& \sum_{n=1}^{2k+1}\frac{1}{n^p}\\ &=& 1+\sum_{i=1}^k\left(\frac{1}{(2i)^p}+\frac{1}{(2i+1)^p}\right)\\ &<&1+\sum_{i=1}^k\frac{2}{(2i)^p}\\ &=&1+2^{1-p}S_k\\ &<&1+2^{1-p}S_{2k+1}\;. \end{eqnarray}$$

Then solving for $S_{2k+1}$ yields

$$S_{2k+1}<\frac{1}{1-2^{1-p}}\;,$$

and since the sequence of partial sums is monotonically increasing and bounded from above, it converges.

(See also: Teresa Cohen & William J. Knight, Convergence and Divergence of $\sum_{n=1}^{\infty} 1/n^p$, Mathematics Magazine, 52(3), 1979, p.178. https://doi.org/10.1080/0025570X.1979.11976778)

My personal favourite is a variant of a common proof that the harmonic series diverges: we get $$\sum_{n=2^k}^{2^{k+1}-1}\frac1{n^p}\le2^k\cdot\frac1{2^{kp}}=2^{(1-p)k}.$$ because the sum has $2^k$ terms of which the first is the largest. Now sum this over $k$ to get $$\sum_{n=1}^{\infty}\frac1{n^p}\le\sum_{k=0}^\infty2^{(1-p)k}=\frac1{1-2^{1-p}}<\infty$$ since $2^{1-p}<1$.

FWIW, the properties of the generalized harmonic series $\sum \frac{1}{n^p}$ can be used to prove another convergence criterion (which goes under the name of second Cauchy's convergence criterion, as far as I remember).

The criterion is the following:

Let $(a_n)$ be a sequence of positive numbers. If:

$$\lim_{n\to \infty} \frac{\ln \frac{1}{a_n}}{\ln n} =L>1 \tag{1}$$

then the series $\sum a_n$ converges. On the other hand, if:

$$\lim_{n\to \infty} \frac{\ln \frac{1}{a_n}}{\ln n} =l<1 \tag{2}$$

then the series $\sum a_n$ diverges.

The proof is very simple. The criterion remains valid even if one replaces $\displaystyle \lim_{n\to \infty}$ with $\displaystyle \liminf_{n\to \infty}$ in (1) and with $\displaystyle \limsup_{n\to \infty}$ in (2).