A Simple Proof That the p-Series Converges for p > 1

Proving that the p-series \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p} converges for p > 1 is a standard exercise in second-semester calculus.  It’s also an important property to know when establishing the convergence of other series via a comparison test.  The usual way I do this in class is with the integral test.  However, I saw a different, simple proof by joriki on Math.SE a few years back.  That proof is the subject of this post.

Let S_n be the nth partial sum \displaystyle S_n = \sum_{k=1}^n \frac{1}{k^p}.  Clearly, \{S_n\} is increasing.  We have

\displaystyle S_{2n+1} = \sum_{k=1}^{2n+1} \frac{1}{k^p} = 1 + \sum_{i=1}^n \left( \frac{1}{(2i)^p} + \frac{1}{(2i+1)^p} \right) < 1 + \sum_{i=1}^n \frac{2}{(2i)^p}

\displaystyle = 1 + 2^{1-p} S_n < 1 + 2^{1-p} S_{2n+1}.

Solving for S_{2n+1} yields (provided p > 1)

\displaystyle S_{2n+1} < \frac{1}{1-2^{1-p}}.

(Again, this only holds when p > 1.  If p = 1 then you can’t isolate S_{2n+1}.  If p < 1 then 1 - 2^{1-p} < 0, and so we get S_{2n+1} > \frac{1}{1-2^{1-p}}.)

Thus the sequence \{S_n\} is bounded above for p > 1.  By the Monotone Convergence Theorem, then, \{S_n\} converges.  And, of course, since the sequence of partial sums converges, \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p} converges.

One use of this proof, as a commenter on joriki’s post mentions, is when doing the series material at the beginning of a Calculus II course rather than at the end.  The only piece of content from the rest of Calculus II that you need for the series material is the integral test, and the primary use of the integral test is to prove convergence of the p-series when p > 1.  With this proof one can establish this convergence without knowing improper integration or the integral test.

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2 Responses to A Simple Proof That the p-Series Converges for p > 1

  1. LNT says:

    Your proof doesn’t use the condition that p>1, i.e. it “proves” that ALL p-series converge, which is clearly not true.

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