Sum of infinite series, Are both of these series equal? 1/2+1/3+1/4...

Question

from an old Numberphile video they explain that the sum of all natural numbers is equal to -1/12, 1+2+3+4+5+...= -1/12. Obviously it diverges, but the -1/12 is meant to be a meaningful representation of that infinite series.

In the extra footage for that video they also say that the -1/12 can be derived in a lot of ways including through the Reimann Zeta function.

When I was looking at the Reimann Zeta function it looked like they are summing 1 + 1/2 + 1/3 + 1/4 + 1/5..... Additionally, I also know that this harmonic series also diverges.

My question is, does 1 + 1/2 + 1/3 + 1/4 + 1/5..... also equal to -1/12 in the same way that the sum of all natural numbers does?

Please let me know.

Thank you.

Answer 1

The sum over all natural numbers actually diverges. What is true is that the Riemann zeta function $\zeta(s)$, which is defined by

$$ \zeta(s)=\sum_{n\ge1}{1\over n^s} $$

for $\Re(s)>1$, can be extended to $\mathbb C\setminus\{1\}$ via analytic continuation. The $-\frac1{12}$ is just what you get when plugging $s=-1$ into the extended Riemann zeta function.

Since the harmonic series diverges at $s=1$, it is impossible to assign $-{1\over12}$ as in the case of natural numbers.

Although $\zeta(s)$ diverges at $s=1$, it is still possible to find its Cauchy principle value at $s=1$, thus allowing us to still assign something for the divergent harmonic series. For convenience, we directly quote some classical results from analytic number theory:

Lemma 1: Let $a_n$ be a complex sequence, and $$A(x)=\sum_{n\le x}a_n.$$ Then we have $$\sum_{n\ge1}{a_n\over n^s}=s\int_0^\infty A(e^t)e^{-st}\mathrm dt.$$ Lemma 2: For $x\ge1$, we have $$H(x)=\sum_{n\le x}\frac1n=\log x+\gamma+O\left(\frac1x\right)$$, where $\gamma$ is the Euler-Mascheroni constant.

By lemma 1, we get

$$ \zeta(1+\delta)=\sum_{n\ge1}{1\over n^\delta}\cdot\frac1n=\delta\int_0^\infty H(e^t)e^{-\delta t}\mathrm dx $$

Now, plug in lemma 2, we see that when $\Re(\delta)>0$,

\begin{align} \zeta(1+\delta) &=\delta\int_0^\infty(t+\gamma)e^{-\delta t}\mathrm dt+O\left(\delta\int_0^\infty e^{-(\delta+1)t}\mathrm dt\right) \\ &=\delta\int_0^\infty te^{-\delta t}\mathrm dt+\gamma+O(\delta). \end{align}

Although the remaining integral converges only for $\Re(\delta)>0$, we can perform analytic continuation via Laplace transform. As a consequence, we obtain an expansion for $\zeta(1+\delta)$ valid for $|\delta|<1$:

$$ \zeta(1+\delta)=\frac1\delta+\gamma+O(\delta) $$

This suggests that the Cauchy principal value of $\zeta(s)$ at $s=1$ is exactly $\gamma$:

$$ \lim_{\delta\to0}{\zeta(1+\delta)+\zeta(1-\delta)\over2}=\gamma $$

Therefore, you can informally assign $\gamma$ to be the "sum" of reciprocals of positive integers.