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

To prove the convergence of the p-series

$$\sum_{n=1}^{\infty} \frac1{n^p}$$

for $p > 1$, one typically appeals to either the Integral Test or the Cauchy Condensation Test.

I am wondering if there is a self-contained proof that this series converges which does not rely on either test.

I suspect that any proof would have to use the ideas behind one of these two tests.

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).

This is the most direct and elementary way I know how to prove the result, although it only works for powers in the range $[0,1] \cup [2,\infty)$ which is exactly the uninteresting set, and Joriki's answer is much better regardless. I had already written this, and perhaps somebody finds it useful.

First we consider $p=1$. Set $x_n$ equal to the greatest power of 2 less than $\frac{1}{n}$. That is, $$(x_n) = (\frac{1}{2}, \frac{1}{4},\frac{1}{4}, \frac{1}{8}, \frac{1}{8},\frac{1}{8},\frac{1}{8},\dots).$$

Note that $$\sum_{i=1}^{\infty} x_n = \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1}x_j\Big) = \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1} \frac{1}{2^{i+1}}\Big) = \sum_{i=1}^{\infty} \frac{1}{2}, $$ which diverges, and that $x_n < \frac{1}{n}$, which proves $\sum_{n=1}^{\infty} \frac{1}{n}$ also diverges.

If $p = 2$, let $x_n = \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ if $n>1$ and 1 if $n=1$.

Observe that $$\sum_{n=1}^{\infty} x_n = 1 + \sum_{n=2}^{\infty}\Big(\frac{1}{n-1} - \frac{1}{n}\Big) = 1 + 1 - \lim_{n\rightarrow\infty}(1/n) = 2,$$ and so in particular this series converges.

Now $x_n > \frac{1}{n \cdot n} = \frac{1}{n^2},$ so the series $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ converges as well.

Finally if $p>2$ then since $\frac{1}{n^p} < \frac{1}{n^2}$, the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges, and if $0<p<1$ we have $\frac{1}{n^p} > \frac{1}{n}$, and so $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges.