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.

For every $p>1$, one can reduce the problem to the convergence of some geometric series. Then, either one takes the latter for granted or one proves it by an elementary argument. More details below.
Let $N(k)$ denote the integer part of $a^k$, where the real number $a>1$ is defined by $a^{p-1}=2$. The sum of $1/n^p$ over $n$ between $N(k)$ and $N(k+1)$ is at most the number of terms $N(k+1)-N(k)$ times the greatest term $1/N(k)^p$. This contribution behaves like 
$$
(a^{k+1}-a^k)/a^{kp}\sim(a-1)a^{k(1-p)}=(a-1)2^{-k}.
$$
The geometric series of general term $2^{-k}$ converges hence the original series converges.
There is a variant of this, where one considers the slabs of integers limited by the powers of $2$. The contribution of slab $k$ is then at most the number of terms $2^k$ times the greatest term $1/2^{kp}$. This reduces the problem to the convergence of the geometric series of general term $b^k$, where $b=2^{1-p}<1$.
Finally, as everybody knows, a simple proof of the convergence of the geometric series of general term $b^k$, for every $b$ in $(0,1)$ say, is to note that the partial sums $s_k$ are such that $s_0=1$ and $s_{k+1}=1+bs_k$, and to show by induction on $k$ that $s_k\le1/(1-b)$ for every $k\ge0$.
Edit The so-called variant above shows, using $b=2^{1-p}$, that the sum of the series is at most $1/(1-b)=2^p/(2^p-2)$. But the contribution of slab $k$ is also at least the number of terms $2^k$ times the smallest term $1/2^{(k+1)p}$, hence the sum of the series is at least $1/(2^p-2)$. Finally the sum of the series is between $1/(2^p-2)$ and $2^p/(2^p-2)$.
