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.

Eight years after I posted this, Zach Teitler pointed out that this proof was already published in $1979$ by Teresa Cohen and William J. Knight in Convergence and Divergence of $\sum_{n=1}^{\infty} 1/n^p$, Mathematics Magazine, $52(3)$, $1979$, p. $178$ (accessible version, DOI link).
