2

I happen to read this post about trying to get the formula of $\sum_{n=1}^N n^m$ for Perron's formula. The general Perron's formula is

$$\sum'_{n\le x} a(n)=\frac{1}{2\pi i}\int_{\text{Re }s=c}f(s)x^s s^{-1}ds$$

where the prime on the summation indicates that the last term of the sum must be multiplied by 1/2 when x is an integer.

I made the following "ill proof" and want to know where I get it wrong:

Now let $x=N+\epsilon$ for $0<\epsilon <1$, then we can ignore the prime on the summation and write

$$\sum_{n\le N} a(n)=\frac{1}{2\pi i}\int_{\text{Re }s=c}f(s)(N+\epsilon)^s s^{-1}ds$$

Let $a(n)=n^m$, then $f(s)=\zeta(s-m)$. Now the formula above says

$$\sum_{n\le N} n^m=\frac{1}{2\pi i}\int_{\text{Re }s=c}\zeta(s-m)(N+\epsilon)^s s^{-1}ds$$

for $c>m+1$. Shifting the line of integration from $s>m+1$ to $s=1$, the only pole encountered is $s=m+1$ and the residue theorem gives us

$$\sum_{n\le N} n^m=\frac{(N+\epsilon)^{m+1}}{m+1}+\frac{1}{2\pi i}\int_{\text{Re }s=1}\zeta(s-m)(N+\epsilon)^s s^{-1}ds$$

The second term is $O(N)$ as $N$ goes to infinity because $\zeta$ is uniformly bounded in a vertical strip near $1-m$. So for high degree (>=2) terms, the coefficients of the binomial expansion of $\frac{(N+\epsilon)^{m+1}}{m+1}$ shall agree with the expression of $\sum_{n\le N} n^m$, in polynomial of $N$.

Now it turns out this is not true, because the coefficients are dependent on the choice of $\epsilon$.

Where is my mistake?

Thanks!

Community
  • 1
AlgRev
  • 583

1 Answers1

3

Excellent question! The problem is that it is not true that the shifted integral

$$ \int_{(1)} \zeta(s-m) (N+\epsilon)^s \frac{ds}{s}$$

converges. The reason for this is that $$ \zeta(\sigma + it) \approx \lvert t \rvert^{\mu(\sigma)}$$ where $$ \begin{cases} \mu(\sigma) = 0 & \text{ if } \sigma \geq 1 \\ \mu(\sigma) = 1/2 - \sigma & \text{ if } \sigma \leq 0 \\ \mu(\sigma) = ? & \text{ if } 0 < \sigma < 1 \end{cases}.$$ (In the middle of the strip the behavior is a bit more nuanced, and this is where the "convexity bound" comes into play). This observation comes directly from the functional equation $$ \pi^{-s/2} \Gamma(s/2)\zeta(s) = \Lambda(s) = \Lambda(1-s)$$ and applying Stirling's approximation to the Gamma functions. In particular, for $\mathrm{Re}(s) > 1$, it's clear that $\zeta(s)$ is uniformly bounded in vertical strips, and so for $\mathrm{Re}(s) < 0$ and Stirling's approximation, we can get very accurate descriptions of the growth of $\zeta(s)$ in terms of a ratio of Gamma functions.

Here, this means that your integral resembles $$\int_{(1)} \lvert t \rvert^{m - \frac{1}{2}} (N+\epsilon)^s \frac{ds}{s} \sim \int_{(1)} \frac{\lvert s \rvert^{m - \frac{1}{2}}}{\lvert s \rvert} (N+\epsilon)^s ds,$$ which diverges. If you add an additional damping term, then you can carry out this argument completely. In fact, this is essentially what Terry Tao used as inspiration for his blog post on Bernoulli numbers, the zeta function, and analytic continuation (which is an excellent read).

Daniel Fischer
  • 206,697
davidlowryduda
  • 91,687