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 $p &gt; 1$, 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:

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

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

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

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

(See also: Teresa Cohen &amp; 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)
Let $S(n)$ = $ \Sigma_{1}^{n} \frac{1}{n^p}$

then 

$S(2n)$ = $S(n)$ + $\frac{1}{{(n+1)}^p} $ +  $\frac{1}{{(n+2)}^p} $ + $\frac{1}{{(n+3)}^p} $ +$\frac{1}{{(n+4)}^p} $ ...........+ $\frac{1}{{(n+n)}^p} $

Let $\Delta S$ = $S(2n) - S(n)$ 

then 
$\frac{n}{{(2n)}^p}$  $\leq$ $\Delta S$ $\leq$  $\frac{n}{(n+1)^p} $

By sandwitch theorem we see that if $p &gt; 1$,  $\lim_{n \rightarrow \inf}$ $\Delta S = 0$

so that as n tends to infinity  $S(n) = S(2n) = S(4n) ....$

So that series converges

EDIT:

$S(2n) - S(n) \leq  \frac{n}{(n+1)^p}  \leq n^{1-p} $

$S(2^{k+1}n) - S(2^{k}n) \leq  \frac{2^{k+1}n - 2^{k}n}{(2^kn+1)^p} \leq \frac{2^{k+1}n - 2^{k}n}{(2^kn)^p} \leq n^{1-p}2^{k(1-p)}$

Summing above equation for k = 0 to m, we get

$S(2^{m+1}n) - S(n) \leq   n^{1-p}(\frac{1-2^{(m+1)(1-p)}}{1-2^{(1-p)}})$

Hence $\lim_{n \to \infty} R(n) =\lim_{m \to \infty}S(2^{m+1}n) - S(n) =0 $