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Consider the Dirichlet series $\sum_{n=1}^{\infty} n^{-s}$, with $s=\sigma+it$, $\sigma$ and $t$ real. How can one prove that this series diverges for $\sigma=1$ and $t\neq 0$?

In all the other combinations of values of $\sigma$ and $t$, it is rather easy to determine (and prove) whether the series converges or diverges:

  1. In the case $\sigma=1$ and $t=0$, the Dirichlet series becomes the harmonic series, which, of course, diverges.

  2. In the case $\sigma>1$ (and any $t$), it is trivial to prove, by the integral test, that the series converges absolutely.

  3. The case $\sigma<1$ (and any $t$) is only slightly more complicated than the previous ones. In this case it is to be proven that the series diverges. To this end one uses a version of the Euler-Maclaurin formula, namely \begin{equation} \sum_{n=1}^{N}f(n) - f(1) = \int_{1}^{N}f(x)dx+\int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx, \end{equation} where $\lfloor x \rfloor$ is the floor function (returning the greatest integer less than or equal to $x$). By playing with the triangle inequality one gets \begin{equation} \left| \sum_{n=1}^{N}f(n) \right| \geq \left| \int_{1}^{N}f(x)dx\right| - \left| \int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx \right| - \left|f(1)\right|; \end{equation} Now notice that $\left| \int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx \right|\leq \int_{1}^{N}\left|f'(x)\right|dx$, and that for the relevant function $f$ (namely, $f(x)=x^{-s}$), the latter converges as $N\to +\infty$. However, in the same limit, $\left| \int_{1}^{N}f(x)dx\right| $ diverges to infinity; from the formula above, it then follows that the Dirichlet series itself diverges to infinity.

The difficulty when $\sigma=1$, $t\neq 0$ is that $\left| \int_{1}^{N}f(x)dx\right| $ does not diverge to infinity; instead, we have $\int_{1}^{N}1/x^{1+it}dx=\frac{i}{t}\left(-1+e^{-it\ln N}\right)$, which---in the $N\to \infty$ limit---has an oscillatory divergence. Therefore I suppose the series $\sum_{n=1}^{\infty} n^{-s}$ should also diverge in an oscillatory manner; however, I don't know how to use the second integral, $\int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx$, in order to prove this.

Perhaps the proof should not use Euler-Maclaurin at all; I just don't know.

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You are right that it oscillates without diverging: as $X\to \infty$, $\sum_{n<X}n^{-1-it} = \zeta(1+it)-X^{-it}/it+o(1)$. –  David Hansen May 21 '11 at 8:00
    
I'd guess that summation by parts has more to offer. You have a harmonic series with coefficients of absolute value 1; so the normal approach would be to take the differences of the reciprocals, getting an absolutely convergent series, and the sum of the factors of absolute value 1. The amount of cancellation in the sum is controlling things, really; and this is a basic type of exponential sum. Not that it is an easy thing, but the old books on the zeta-function would be a place to look for insight. –  Charles Matthews May 21 '11 at 12:10
    
This series can be written as $\sum_{n \geq 1} e^{-it \log n}/n$. The point is that $\log n$ grows slowly as $n \to +\infty$, so sums of the type $\sum_{n=A}^{B} \cdot$ are unbounded and by Cauchy's criterion the series diverges. To see how it diverges I would try Abel summation (which amouns to integration by parts using Stieltjes formalism). –  François Brunault May 21 '11 at 12:11

2 Answers 2

up vote 11 down vote accepted

Write $$\sum_{n=1}^{N} \frac{1}{n^s} = \sum_{n=1}^{N} \left( \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right) + \int_{x=1}^{N+1} \frac{dx}{x^s} \quad (*)$$

You want to know whether the limit of the left hand side of $(*)$ exists, as $N \to \infty$. Now, you can check that $$\left| \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right| \leq \frac{|s|}{n^{\mathrm{Re}(s)+1}}.$$

So $$\sum_{n=1}^{\infty} \left( \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right)$$ is absolutely convergent and the limit of the first term on the right hand side of $(*)$ exists. So the question reduces to whether or not $$\lim_{N \to \infty} \int_{x=1}^{N+1} \frac{dx}{x^s}$$ exists. We can compute the integral explicitly, and see that it does not.

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On reading your question more carefully, this is the same approach you were trying. In your language, note that $|f'(x) (x - \lfloor x \rfloor)| \leq 1/|x|^{s+1}$, so the integral you didn't know how to deal with converges. –  David Speyer May 21 '11 at 13:00
    
Thank you---it seems I somehow managed to forget that if the improper integral of $|g|$ converges, then (by the comparison test for improper integrals) so does the integral of $g$ ! –  retqr May 21 '11 at 18:52

I wanted to correct a couple of the inequalities in David Speyer's answer and comments.

I believe one in the answer should read $\left| \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right| \leq |s|/(2n^{{Re}(s)+1})$. (This has now been corrected by Mr. Speyer, omitting the "2" in the denominator, which is fine.)

Also, I believe the one in the following comment should read $|f'(x) (x - \lfloor x \rfloor)| \leq |s|/|x|^{{Re}(s)+1}$.

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Thanks! The answer is fixed; not bothering with the comment. –  David Speyer Dec 6 '11 at 11:36

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