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.
Best Answer
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)