In one of the letters to G. H. Hardy, Srinivasa Ramanujan wrote that "I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal.".
So according to Ramanujan's theory, we get the following formula, which is remarkable and seemingly impossible.
$$\sum_{n=1}^{\infty}n=1+2+3+\cdots=-\frac{1}{12}$$
When I first came to know about this formula, I was astounded. How can a divergent sum can possibly have a finite value, and also a negative one? Over past few years, I occasionally tried to understand the meaning behind this equation, with no success. Today I decided to give it a try again. I still can't fully grasp it, but I shall try my best to explain what little I understood so far.
The formula is completely incorrect if it is interpreted in traditional manner. In last post I discussed briefly about zeta function. It is defined as $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$
Zeta function has finite values for $\text{Re}(s)>1$. For $\text{Re}(s)<=1$ zeta function becomes divergent. Zeta function is an analytical function, i.e. on a domain $D$ if it is complex differentiable at every point in $D$. In complex analysis there is a method known as 'Analytic Continuation' which extend the domain of an analytical function. I am not familiar with such method, so I shall use a method known as 'Cutoff regularization' which is little bit easier to comprehend.
A cutoff function $f$ is defined as $f:\textbf{R}^+ \to \textbf{R}$ which equals $1$ at $0$. We truncate our target sum at $N$. So we need to evaluate $\sum_{n=1}^{N}n$. After including the cutoff regularization, we get
$$\sum_{n=1}^{N}nf\left(\frac{n}{N}\right)$$
As cutoff function takes $1$ at $0$, so we get back our original sum when $N \to \infty$, i.e. we have to evaluate $$\sum_{n=1}^{\infty}nf\left(\frac{n}{N}\right)$$
By introducing a cutoff function, we replaced partial sums with smoothed sums. Using any cutoff function which follow above condition, we can show that the value obtained by analytic continuation is identical to the constant term of the smoothed sums. Some really heavy machinery is required to prove this and I got completely lost here. But we can easily verify it using an example.
Let $f(n/N)=e^{-n/N}$. $f(n/N)=1$ when $n \to 0$, i.e. $N \to \infty$. Now we shall try to evaluate the smoothed sum. Let $\frac{1}{N}=\theta$
$$\begin{array}{lcl}\sum_{n=1}^{\infty} ne^{-\theta n} & = & - \frac{d}{d\theta}\sum_{n=1}^{\infty} e^{-\theta n}\\
& = & - \frac{d}{d\theta} \left( \frac{1}{1-e^{-\theta}} \right)\\
& = & \frac{e^{-\theta}}{(1-e^{-\theta})^2}\\
& = & \frac{1}{\theta^2}-\frac{1}{12}+\frac{\theta^2}{240} - \frac{\theta^4}{6048} + \frac{\theta^6}{172800}- \frac{\theta^8}{5322240} + O(\theta^9)\end{array}$$
So we got the fabled analytic expansion value $-\frac{1}{12}$ as the constant term of smoothed sum.
$$\zeta(-1)=-\frac{1}{12}$$
Using similar method to obtain the value of zeta function at other negative integers.
Analytic continuation values can be interpreted geometrically. Partial sum of first $N$ natural numbers is given by $\frac{N(N+1)}{2}$. Let $g(x)=\frac{x(x+1)}{2}$, which extends the discrete partial terms to continuous domain. I plotted the function for $x=-1.5$ to $x=1.5$.
$g(x)$ cuts x-axis at $x=-1$ and $x=0$. So the small area under x-axis is given by
$$\int_{-1}^{0} \frac{x(x+1)}{2} dx=-\frac{1}{12}$$
We got the analytic continuation value again !!
We can obtain the analytic continuation value for other negative integers in similar fashion. Due to the odd symmetry, for all negative even integers, zeta function takes the value $0$. In general
$$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$ where $B_n$ is the Bernoulli numbers, which are defined as $$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_nx^n}{n!}$$
In future, I shall try to discuss about it in detail if I manage to understand the underlying connection between analytic expansion and this geometrical method.
Thanks for reading :)
So according to Ramanujan's theory, we get the following formula, which is remarkable and seemingly impossible.
$$\sum_{n=1}^{\infty}n=1+2+3+\cdots=-\frac{1}{12}$$
When I first came to know about this formula, I was astounded. How can a divergent sum can possibly have a finite value, and also a negative one? Over past few years, I occasionally tried to understand the meaning behind this equation, with no success. Today I decided to give it a try again. I still can't fully grasp it, but I shall try my best to explain what little I understood so far.
The formula is completely incorrect if it is interpreted in traditional manner. In last post I discussed briefly about zeta function. It is defined as $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$
Zeta function has finite values for $\text{Re}(s)>1$. For $\text{Re}(s)<=1$ zeta function becomes divergent. Zeta function is an analytical function, i.e. on a domain $D$ if it is complex differentiable at every point in $D$. In complex analysis there is a method known as 'Analytic Continuation' which extend the domain of an analytical function. I am not familiar with such method, so I shall use a method known as 'Cutoff regularization' which is little bit easier to comprehend.
A cutoff function $f$ is defined as $f:\textbf{R}^+ \to \textbf{R}$ which equals $1$ at $0$. We truncate our target sum at $N$. So we need to evaluate $\sum_{n=1}^{N}n$. After including the cutoff regularization, we get
$$\sum_{n=1}^{N}nf\left(\frac{n}{N}\right)$$
As cutoff function takes $1$ at $0$, so we get back our original sum when $N \to \infty$, i.e. we have to evaluate $$\sum_{n=1}^{\infty}nf\left(\frac{n}{N}\right)$$
By introducing a cutoff function, we replaced partial sums with smoothed sums. Using any cutoff function which follow above condition, we can show that the value obtained by analytic continuation is identical to the constant term of the smoothed sums. Some really heavy machinery is required to prove this and I got completely lost here. But we can easily verify it using an example.
Let $f(n/N)=e^{-n/N}$. $f(n/N)=1$ when $n \to 0$, i.e. $N \to \infty$. Now we shall try to evaluate the smoothed sum. Let $\frac{1}{N}=\theta$
$$\begin{array}{lcl}\sum_{n=1}^{\infty} ne^{-\theta n} & = & - \frac{d}{d\theta}\sum_{n=1}^{\infty} e^{-\theta n}\\
& = & - \frac{d}{d\theta} \left( \frac{1}{1-e^{-\theta}} \right)\\
& = & \frac{e^{-\theta}}{(1-e^{-\theta})^2}\\
& = & \frac{1}{\theta^2}-\frac{1}{12}+\frac{\theta^2}{240} - \frac{\theta^4}{6048} + \frac{\theta^6}{172800}- \frac{\theta^8}{5322240} + O(\theta^9)\end{array}$$
So we got the fabled analytic expansion value $-\frac{1}{12}$ as the constant term of smoothed sum.
$$\zeta(-1)=-\frac{1}{12}$$
Using similar method to obtain the value of zeta function at other negative integers.
Analytic continuation values can be interpreted geometrically. Partial sum of first $N$ natural numbers is given by $\frac{N(N+1)}{2}$. Let $g(x)=\frac{x(x+1)}{2}$, which extends the discrete partial terms to continuous domain. I plotted the function for $x=-1.5$ to $x=1.5$.
$g(x)$ cuts x-axis at $x=-1$ and $x=0$. So the small area under x-axis is given by
$$\int_{-1}^{0} \frac{x(x+1)}{2} dx=-\frac{1}{12}$$
We got the analytic continuation value again !!
We can obtain the analytic continuation value for other negative integers in similar fashion. Due to the odd symmetry, for all negative even integers, zeta function takes the value $0$. In general
$$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$ where $B_n$ is the Bernoulli numbers, which are defined as $$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_nx^n}{n!}$$
In future, I shall try to discuss about it in detail if I manage to understand the underlying connection between analytic expansion and this geometrical method.
Thanks for reading :)
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