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

n=11np

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.

Answer

We can bound the partial sums by multiples of themselves:

S2k+1=2k+1n=11np=1+ki=1(1(2i)p+1(2i+1)p)<1+ki=12(2i)p=1+21pSk<1+21pS2k+1.

Then solving for S2k+1 yields

S2k+1<1121p,

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, Mathematics Magazine, 52(3), 1979, p.178. https://doi.org/10.1080/0025570X.1979.11976778)

Attribution
Source : Link , Question Author : admchrch , Answer Author : Zach Teitler

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