How does one get that $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$?

I think I have rediscovered it (after watching the video you linked and read wiki biography of Ramanujan).

Start with $$ \frac{1}{x+1} =1 -x +x^2 -x^3 +-\cdots, \quad |x| <1. $$ and differentiate to get $$ -\frac{1}{(x+1)^2} =-1 +2x -3x^2 +4x^3 -+\cdots, \quad |x| <1. \\ \frac{2}{(x+1)^3} =2 \cdot 1 -3 \cdot 2x +4 \cdot 3 x^2 -+\cdots, \quad |x| <1. \\ -\frac{6}{(x+1)^4} =-3 \cdot 2 \cdot 1 +4 \cdot 3 \cdot 2x -5 \cdot 4 \cdot 3 x^2 +-\cdots, \quad |x| <1. $$

Take a magic mushroom and, ignoring $|x| <1$, let us take $x=1$ in each one. $$ \frac{1}{2} =1 -1 +1 -1 +- \cdots \\ -\frac{1}{4} =-1 +2 -3 +4 -+ \cdots \\ \frac{1}{4} =2 \cdot 1 -3 \cdot 2 +4 \cdot 3 -+ \cdots \\ -\frac{3}{8} =-3 \cdot 2 \cdot 1 +4 \cdot 3 \cdot 2 -5 \cdot 4 \cdot 3 +- \cdots $$ Or more formally, $$ \sum_{m=1}^\infty (-1)^{m+1} m =\frac{1}{4}. \\ \sum_{m=1}^\infty (-1)^{m+1} m (m+1) =\frac{1}{4}. \\ \sum_{m=1}^\infty (-1)^{m+1} m (m+1) (m+2) =\frac{3}{8}. $$

But notice $$ \begin{align} \sum_{m=1}^\infty (-1)^{m+1} m^2 =&\sum_{m=1}^\infty (-1)^{m+1} m (m+1) -\sum_{m=1}^\infty (-1)^{m+1} m \\ =&\frac{1}{4} -\frac{1}{4} =0. \end{align} $$ and $$ \begin{align} \sum_{m=1}^\infty (-1)^{m+1} m^3 =&\sum_{m=1}^\infty (-1)^{m+1} m (m+1) (m+2) -3\sum_{m=1}^\infty (-1)^{m+1} m^2 -2\sum_{m=1}^\infty (-1)^{m+1} m \\ =&\frac{3}{8} -3 \cdot 0 -2 \cdot \frac{1}{4} =-\frac{1}{8} \quad \quad \ldots \spadesuit \end{align} $$

On the other hand, $$ \zeta(-3) :=1^3 +2^3 +3^3 +\cdots \\ 2^4 \zeta(-3) =2 \cdot 2^3 +2 \cdot 4^3 +2 \cdot 6^3 +\cdots \\ $$ Subtract them, aligning the 2nd, 4th, 6th term like Ramanjunan did in his notebooks (shown in the video). $$ -15 \zeta(-3) =1^3 -2^3 +3^3 -+\cdots \quad \quad \ldots \heartsuit $$ $\heartsuit$ and $\spadesuit$ together give us: (Hold your breath.) $$ \sum_{m=1}^\infty m^3 =\frac{1}{120}. $$

Recently I also found a proof of Riemann conjecture, but the answer box is too narrow for me to type all that down.

P.s. seriously, I think Ramanujan's effort is sort of finding an interpretation of divergent series so that they have a real value, while their manipulation to be still consistent to our usual notion of arithmetics: arranging, addition, expanding, etc.? Maybe this can be compared to the attempt to define quaternion as an extension of complex numbers, while inevitably discarding commutative law?

Let $f(x)=x^d$ for $d\geq 1$ then by Euler–Maclaurin summation formula $$\sum_{k=1}^nf(x)=C+\int_0^n f(x) dx+\frac{f(n)}{2}+\sum_{k\geq 1}\frac{B_{2k}}{(2k)!}f^{(2k-1)}(n)$$ where $B_n$ denotes the $n$-th Bernoulli number.

In his theory of Divergent Series, Ramanujan "identified" $\sum_{k=1}^{\infty}f(x)$ with $C$. Note that $C$ is zero if $d$ is even and it is $-\frac{B_{d+1}}{d+1}$ if $d$ is odd. So for $d=1$ we get $-\frac{1}{12}$ and for $d=3$ we obtain $\frac{1}{120}$.

Below is the derivation of the following equality:

$$\zeta(-s)=\frac1{1-2^{1+s}}\lim_{r\to1^-}\left(\underbrace{r\cdot\frac d{dr}r\cdot\frac d{dr}\dots r\cdot\frac d{dr}}_s\frac{-1}{1+r}\right)$$

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Notice that if we have

$$\zeta(s)=1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\dots$$

$$\eta(s)=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\dots$$

then, when they are absolutely convergent, we get the functional equation:

$$\begin{align}\zeta(s)&=1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\dots\\\eta(s)&=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\dots\\\hline\zeta(s)-\eta(s)&=2\left(\ \ \frac1{2^s}\qquad\ \ +\frac1{4^s}+\dots\right)\\&=2^{1-s}\left(1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\dots\right)\tag{factored out $2^s$}\\&=2^{1-s}\zeta(s)\end{align}$$

Solving for $\zeta(s)$, we get

$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$

And since these are analytic functions, this holds for all $s$.

As a somewhat elementary method, we can evaluate the $\eta(s)$ as follows, which converges for all $s$.

$$\eta(s)=\lim_{r\to1^-}\sum_{n=1}^\infty\frac{(-1)^{n+1}r^n}{n^s}$$

And thus, we have ourselves left with

$$\eta(-3)=\lim_{r\to1^-}\sum_{n=1}^\infty(-1)^{n+1}r^nn^3=-\frac18$$

which may be done by manipulating the geometric series:

$$\frac{-1}{1+r}=\sum_{n=0}^\infty(-1)^{n+1}r^n$$

$$r\cdot\frac d{dr}\frac{-1}{1+r}=\sum_{n=0}^\infty(-1)^{n+1}nr^n\\\vdots$$

$$\underbrace{r\cdot\frac d{dr}r\cdot\frac d{dr}\dots r\cdot\frac d{dr}}_k\frac{-1}{1+r}=\sum_{n=0}^\infty(-1)^{n+1}n^kr^n$$

Which then yields the desired result $\zeta(-3)=\frac1{120}$.

So, succinctly, we end up with the following form for the zeta function whenever $s\in\mathbb N,\ s>0$:

$$\zeta(-s)=\frac1{1-2^{1+s}}\lim_{r\to1^-}\left(\underbrace{r\cdot\frac d{dr}r\cdot\frac d{dr}\dots r\cdot\frac d{dr}}_s\frac{-1}{1+r}\right)$$

which isn't terribly difficult to work out. Just a lot of quotient rule going on in there. If $s=0$, we have

$$\zeta(0)=\frac1{1-2^{1+0}}\lim_{r\to1^-}\left(\frac1{1+r}\right)=-\frac12$$

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Some values worked out for you:

$$\zeta(-1)=\frac1{1-2^{1+1}}\lim_{r\to1^-}\left(r\cdot\frac d{dr}\frac{-1}{1+r}\right)=-\frac13\lim_{r\to1^-}\left(\frac{r}{(1+r)^2}\right)=-\frac1{12}$$

$$\zeta(-2)=\frac1{1-2^{1+2}}\lim_{r\to1^-}\left(r\cdot\frac d{dr}r\cdot\frac d{dr}\frac{-1}{1+r}\right)=-\frac17\lim_{r\to1^-}\left(r\cdot\frac d{dr}\frac{r}{(1+r)^2}\right)\\=-\frac17\lim_{r\to1^-}\left(\frac{r-r^2}{(1+r)^3}\right)=0$$

$$\zeta(-3)=\frac1{1-2^{1+3}}\lim_{r\to1^-}\left(r\cdot\frac d{dr}r\cdot\frac d{dr}r\cdot\frac d{dr}\frac{-1}{1+r}\right)=-\frac1{15}\lim_{r\to1^-}\left(r\cdot\frac d{dr}\frac{r-r^2}{(1+r)^3}\right)\\=-\frac1{15}\left(\frac{r-4r^2+r^3}{(1+r)^4}\right)=\frac1{120}$$