TITLE: Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$
QUESTION [178 upvotes]: 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.

REPLY [2 votes]: We first show that for $k\geq 1$
$$
\sum_{n=1}^\infty\frac{1}{n(n+1)\dotsb(n+k)}=\frac{1}{k\times k!}\tag{0}.
$$
This result combined with the limit comparison test will yield the fact that $\sum n^{-p}<\infty$ for $p>1$. 
Indeed, the series in (0) telescopes as can be seen from the partial fraction decomposition
$$
\frac{1}{n(n+1)\dotsb(n+k)}=\frac{1}{k}\left(
\frac{1}{n(n+1)\dotsb(n+k-1)}-\frac{1}{(n+1)\dotsb(n+k)}
\right)
$$
whence
$$
\sum_{n=1}^m\frac{1}{n(n+1)\dotsb(n+k)}=\frac{1}{k}\left(
\frac{1}{k!}-\frac{1}{(m+1)\dotsb(m+k)}
\right).
$$
Letting $m\to\infty$ yields the result. 
Next observe that for $p>1$ we have that
$$
\frac{1}{n^p}\sim\frac{1}{n(n+1)\dotsb(n+p-1)}.
$$
Apply the limit comparison test and we are done.