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.

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 $p > 1$,  $\lim_{n \rightarrow \inf}$ $\Delta S = 0$
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 $
