Clarifying the series

I’m confirming the series in question, likely the famous 1 + 2 + 3 + ... = -1/12, known as Ramanujan's Summation, often shown through zeta function regularization.

Applying the Riemann zeta function

I’m showing how the infinite series S is linked to the Riemann zeta function \zeta(-1) = -1/12, by analytically continuing it to negative s.

Perusing various methods

I’m noting that Ramanujan summation or zeta function regularization yields -1/12. Other methods include the polygamma function and the Dirichlet eta function, though they lack classical rigor.

Thought about Ramanujan's summation of series for 8 seconds

Ramanujan’s summation (or more broadly, zeta-function regularization) is a way to assign finite values to certain divergent infinite series. The most famous example is the assignment

This is not a valid equality in the usual sense of partial sums (which of course diverge). Instead, it is a statement about an analytic continuation of special functions (particularly the Riemann $\zeta$-function) or, equivalently, about a summation method introduced by Ramanujan. Below is an outline of why one may consistently define or regularize the value of $1 + 2 + 3 + \dots$ to be $-\tfrac{1}{12}$.

1. The classical Riemann $\zeta$-function approach

1.1 Definition of the Riemann $\zeta$-function

For $\Re(s) > 1$, the Riemann $\zeta$-function is defined as

In this region, the series converges absolutely. One of the fundamental results in complex analysis is that $\zeta(s)$ can be analytically continued to a meromorphic function on the entire complex plane, except for a simple pole at $s = 1$.

1.2 Analytic continuation and value at $s=-1$

Once we know that $\zeta(s)$ extends (continues) to all complex $s$, it makes sense to evaluate it at negative integers. In particular, at $s = -1$, it turns out that

Mathematically, if we try to interpret

as $\zeta(-1)$, then the analytic continuation gives the value $-\tfrac{1}{12}$. That is the sense in which one says

It’s critical to emphasize that the partial sums $1, \;1+2=3,\;1+2+3=6,\;\dots$ obviously grow without bound. Hence, in the usual sense, the series diverges. The value $-\tfrac{1}{12}$ is not a “sum” in the standard elementary sense but in the regularized or Ramanujan-summed sense.

2. A glimpse into Ramanujan’s method

Ramanujan pioneered another viewpoint to assign finite values to divergent series. While his derivations can look like “formal manipulations,” they are shorthand for more rigorous justifications using analytic continuation or other summation methods. Here is a sketch.

2.1 A simpler series: $1 - 1 + 1 - 1 + \dots$

Consider the series

If you group terms as $(1-1) + (1-1) + \dots$, you get $0$. But if you shift and group as $1 + (-1+1) + (-1+1) + \dots$, you get $1$. The partial sums oscillate: $1, 0, 1, 0, 1, \dots$

A Cesàro or Ramanujan summation approach can assign the average of these partial sums as the summation value. In fact, it is consistent to define

(This agrees with the Dirichlet $\eta$-function at $s=0$, giving $\eta(0)=\frac12$.)

2.2 Another series: $1 - 2 + 3 - 4 + \dots$

A famous Ramanujan manipulation goes as follows:

1. Let $$S_2 = 1 - 2 + 3 - 4 + 5 - 6 + \dots$$
2. Observe that $$(1 - 2 + 3 - 4 + \dots) - (\, -1 + 2 - 3 + 4 - \dots) \;=\; 1 + 1 + 1 + \dots$$ and using various partial summation methods, one finds $$S_2 = \frac{1}{4}.$$ (This also matches the analytic value $\eta(-1) = \tfrac14$.)

2.3 The famous series: $1 + 2 + 3 + 4 + \dots$

Ramanujan used a chain of manipulations (often in his letters) that went somewhat like this:

1. Suppose we define

   $$S = 1 + 2 + 3 + 4 + \dots$$

   Of course, $\sum_{n=1}^N n \to \infty$ as $N\to\infty$. But let’s see how a formal trick emerges:

2. Consider

   $$S_1 \;=\; 1 - 1 + 1 - 1 + \dots = \tfrac12,$$

   and

   $$S_2 \;=\; 1 - 2 + 3 - 4 + 5 - 6 + \dots = \tfrac14.$$

3. Look at

   $$S - 4S_2 \;=\; (1 + 2 + 3 + 4 + \dots) - 4(1 - 2 + 3 - 4 + \dots).$$

   One can attempt to rearrange and group terms in a formal way, eventually deducing $S = -\tfrac{1}{12}$.

What Ramanujan was doing is a heuristic outline of the deeper fact that $\zeta(-1) = -\tfrac{1}{12}$. It requires methods beyond ordinary convergence (like analytic continuation) to be made rigorous.

3. A rigorous outline using analytic continuation

To see more formally why $\zeta(-1) = -\tfrac{1}{12}$, one uses the functional equation and the known values of $\zeta(s)$ at negative integers:

1. Gamma function identity:

   $$\Gamma(s)\,\zeta(s) \;=\; \text{(something that has an analytic continuation everywhere)},$$

   and specifically the functional equation for the zeta function:

   $$\zeta(s) = 2^s \pi^{s-1}\sin\Bigl(\frac{\pi s}{2}\Bigr)\Gamma(1-s)\,\zeta(1-s).$$

2. Evaluate at negative integers: Plug in $s=-1$, and you eventually deduce

   $$\zeta(-1) \;=\; -\tfrac{1}{12}.$$

3. Interpretation: If we define the divergent series $\sum_{n=1}^\infty n$ by regularizing it to be $\zeta(-1)$, then we say

   $$1 + 2 + 3 + 4 + \dots \;=\; \zeta(-1) \;=\; -\tfrac{1}{12}.$$

4. Key takeaway (and a word of caution)

• In standard real analysis, the infinite series $1 + 2 + 3 + \dots$ diverges to infinity. There is no finite sum in the classical sense.
• In Ramanujan summation or zeta-function regularization, we assign the symbol $$1 + 2 + 3 + 4 + \dots \;=\; -\tfrac{1}{12}$$ to mean we are evaluating the analytic continuation $\zeta(s)$ at $s=-1$.

Because of its surprising nature, it often appears in physics (string theory, zeta regularization for quantum fields, etc.) and has become a sort of “headline result” in popular math discussions. It must be remembered, however, that this is a specialized notion of summation that does not reflect ordinary partial sums.

References / Further Reading

1. G. H. Hardy, Divergent Series, 1949.
2. Bruce C. Berndt, Ramanujan’s Notebooks.
3. Titchmarsh, The Theory of the Riemann Zeta-Function.
4. Euler’s approach and Riemann’s approach to the zeta function: many references discuss the continuation to negative integers.

Summary:

• Ramanujan’s summation (and related zeta-function regularization) provides a consistent way to assign a finite value $-\tfrac{1}{12}$ to the divergent series $1 + 2 + 3 + 4 + \dots$.
• The real rigor behind this comes from extending $\zeta(s)$ to the region $\Re(s) \le 1$, where the original series for $\zeta(s)$ no longer converges, but the analytically continued function remains well-defined—and specifically takes the value $-\frac{1}{12}$ at $s=-1$.
• All of this hinges on the idea that we are not talking about the usual limit of partial sums but a regularized notion of summation.

Mapping out the approach

I'm outlining the proof of the formula 1 + 2 + 3 + 4 + ... = -1/12 using the Riemann zeta function's functional equation and Ramanujan summation. Evaluating \zeta(-1) after analytic continuation helps.

Unraveling Bernoulli connections

I'm exploring the relationship between the Riemann zeta function $\zeta(s)$ and Bernoulli numbers to derive the formula, involving the Dirichlet eta function $\eta(s)$.

Utilizing functional equations

Evaluating $\zeta(-1)$ through the Riemann zeta functional equation confirms $\zeta(-1) = -1/12$, aligning with the standard result.

Thought about origin of Ramanujan's formula for 17 seconds

Below is a more detailed outline showing where the famous value

“comes from” in a rigorous (analytic) sense. There are several ways to arrive at the result. We will focus on one of the most straightforward methods: using the analytic continuation of the Riemann $\zeta$-function (and its functional equation). We will end by giving an alternative short derivation via Bernoulli numbers, which is often quoted in textbooks.

1. Recap of the Riemann $\zeta$-function

For $\Re(s) > 1$, the Riemann $\zeta$-function is given by

Although this series diverges for $\Re(s) \le 1$, the function $\zeta(s)$ can be analytically continued to a meromorphic function on the whole complex plane, with a single simple pole at $s=1$.

The “magic” is that once $\zeta(s)$ is continued, we can meaningfully talk about values like $\zeta(-1)$, even though the original series $\sum n$ obviously diverges.

2. Using the functional equation

A standard result in analytic number theory is the functional equation for the Riemann zeta function, which (in one common form) says:

where $\Gamma$ is the Gamma function.

We want to evaluate $\zeta(-1)$. Plug $s=-1$ into the functional equation:

1. Left side: $\zeta(-1)$.

2. Right side:

   $$2^{-1} \,\pi^{-2} \,\sin\!\bigl(-\tfrac{\pi}{2}\bigr) \,\Gamma\bigl(1 - (-1)\bigr) \,\zeta\bigl(1-(-1)\bigr).$$

   Let us compute each piece:

   • $2^{-1} = \tfrac12.$
   • $\pi^{-2} = \tfrac{1}{\pi^2}.$
   • $\sin\!\bigl(-\tfrac{\pi}{2}\bigr) = -1.$
   • $\Gamma(1 - (-1)) = \Gamma(2) = 1! = 1.$
   • $\zeta(1-(-1)) = \zeta(2) = \frac{\pi^2}{6}.$

   Putting these together:

   $$2^{-1} \cdot \pi^{-2} \cdot (-1) \cdot 1 \cdot \frac{\pi^2}{6} \;=\; \frac12 \cdot \frac{1}{\pi^2} \cdot (-1) \cdot \frac{\pi^2}{6} \;=\; -\frac{1}{12}.$$

Hence, from the functional equation:

Interpreting this as a “sum”

Recall how $\zeta(s)$ relates to sums:

• For $\Re(s) > 1$, $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$.
• Formally, if you try $s = -1$, you get $\sum_{n=1}^\infty n$. Clearly, $\sum_{n=1}^\infty n$ diverges in the elementary sense.
• But the analytically continued function takes the value $\zeta(-1) = -\tfrac{1}{12}$.

Thus, the Ramanujan summation (or zeta-function regularization) defines:

It is crucial to emphasize: this does not mean the partial sums $1, 3, 6, 10, \dots$ converge to $-\frac{1}{12}$. They obviously diverge to $+\infty$. Rather, $-\tfrac{1}{12}$ is the value assigned by a regularization method (analytic continuation).

3. Short Bernoulli-number derivation

Another classical formula states that for nonnegative integers $n$,

where $B_{k}$ are the Bernoulli numbers.

In particular, taking $n=1$ gives:

But $B_2 = \tfrac{1}{6}$. Therefore,

This is a standard identity found in many references on the Riemann zeta function and Bernoulli numbers (e.g., Titchmarsh’s “Theory of the Riemann Zeta-Function,” or G. H. Hardy’s “Divergent Series”).

4. Summary

• Core idea: The extension of $\zeta(s)$ to negative integers (where the original series definition no longer converges) yields specific finite values.
• Functional equation or Bernoulli numbers both show that $\zeta(-1) = -\tfrac{1}{12}$.
• Hence we say (in the regularized sense): $$1 + 2 + 3 + 4 + \dots \;=\; -\tfrac{1}{12}.$$

This is where the famous formula “comes from” in a rigorous analytic setting.