# Self-Contained Proof that ∞∑n=11np\sum\limits_{n=1}^{\infty} \frac1{n^p} Converges for p>1p > 1

To prove the convergence of the p-series

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.

We can bound the partial sums by multiples of themselves:

$$S2k+1=2k+1∑n=11np=1+k∑i=1(1(2i)p+1(2i+1)p)<1+k∑i=12(2i)p=1+21−pSk<1+21−pS2k+1.\begin{eqnarray} S_{2k+1} &=& \sum_{n=1}^{2k+1}\frac{1}{n^p}\\ &=& 1+\sum_{i=1}^k\left(\frac{1}{(2i)^p}+\frac{1}{(2i+1)^p}\right)\\ &<&1+\sum_{i=1}^k\frac{2}{(2i)^p}\\ &=&1+2^{1-p}S_k\\ &<&1+2^{1-p}S_{2k+1}\;. \end{eqnarray}$$

Then solving for $$S2k+1S_{2k+1}$$ yields

$$S2k+1<11−21−p,S_{2k+1}<\frac{1}{1-2^{1-p}}\;,$$

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 $$∑∞n=11/np\sum_{n=1}^{\infty} 1/n^p$$, Mathematics Magazine, 52(3), 1979, p.178. https://doi.org/10.1080/0025570X.1979.11976778)