In quantum electrodynamics, the **bare** charge of an **electron** is infinite, but the renormalized **dressed** charge is finite. The bare electron shields itself by polarizing the virtual electron-positron pairs of the nearby **quantum vacuum** to reduce its **coupling** at large distances to

\infty - \infty \stackrel{I}{=} \sqrt{\frac{1}{137}}
in natural units, where the “I” decorating the equals sign denotes an **informal** relationship. **Renormalization** techniques help physicists make sense of divergent sums and integrals used to calculate quantities like **running **coupling “constants” that vary with interaction energy and set the strengths of the fundamental forces.

Corresponding sum and integral diverge together

For a simple example, the **sum** of natural numbers

\sum_{n=0}^N n = 1+2+3+\cdots+N= \frac{N}{2}(N+1)
diverges as N\rightarrow \infty. Similarly, the analogous **integral** of positive real numbers

\int_{0}^X \hspace{-0.8em}dx\ x =\frac{X^2}{2}
diverges as X\rightarrow \infty. To cancel these corresponding discrete and continuous divergences, introduce an exponential **convergence factor** and formally subtract the integral from the sum

\sum_{n=0}^\infty n -\int_{0}^\infty \hspace{-0.8em} dx\, x =\color{red}\lim_{\lambda \rightarrow 0}\color{black}\left( \sum_{n=0}^\infty n\, \color{red}e^{-\lambda n}\color{black}-\int_{0}^\infty \hspace{-0.8em} dx\, x\, \color{red}e^{-\lambda x} \color{black}\right).
To evaluate the cancellation, differentiate the integral with respect to the convergence parameter \lambda>0 to get

\mathcal{I}=\int_{0}^\infty \hspace{-0.8em}dx\ x\, \color{red}e^{-\lambda x}\color{black} =-\frac{d}{d\lambda}\int_{0}^\infty \hspace{-0.8em}dx\ e^{-\lambda x} =-\frac{d}{d\lambda}\left(\frac{e^{-\lambda x}}{-\lambda}\right)\bigg|_{0}^\infty=-\frac{d}{d\lambda}\left(\frac{1}{\lambda}\right)=\frac{1}{\lambda^2}.
Next differentiate the sum with respect to \lambda

\mathcal{S}=\sum_{n=0}^\infty n\,\color{red} e^{-\lambda n}\color{black} = -\frac{d}{d\lambda} \sum_{n=0}^\infty e^{-\lambda n} = -\frac{d}{d\lambda} \sum_{n=0}^\infty \left(e^{-\lambda}\right)^n
and sum the resulting geometric series to find

\mathcal{S}= -\frac{d}{d\lambda} \left( \frac{1}{1-e^{-\lambda}} \right)= \frac{e^{-\lambda}}{\left(1-e^{-\lambda}\right)^2}\color{red}\frac{e^{2\lambda}} {e^{2\lambda}}\color{black}= \frac{e^{\lambda}}{\left(e^{\lambda}-1\right)^2}\color{red}= \frac{e^{\lambda}}{\left(1-e^{\lambda}\right)^2}\color{black},
which is thus an even function of \lambda. Replace the exponentials by their power series expansions

\mathcal{S}=\frac{1+\lambda+\lambda^2/2!+\lambda^3/3!+\cdots}{\left(1+\lambda+\lambda^2/2!+\lambda^3/3!+\cdots - 1 \right)^2}=\frac{1+\lambda+\lambda^2/2+\cdots}{\lambda^2\left(1+\lambda/2+\lambda^2/6+\cdots \right)^2},
expand the denominator square

\begin{array}{ccc}\left(1+\lambda/2+\lambda^2/6+\cdots\right)^2=&+1\hspace{1.2em}+\lambda/2\hspace{0.7em}+\lambda^2/6\hspace{0.2em}\phantom{+\cdots}\\ \rule{0pt}{1.3em}&+\lambda/2\hspace{0.43em}+\lambda^2/4\hspace{0.37em}\color{red}+\lambda^3/12\color{black}\phantom{+\cdots}\\ \rule{0pt}{1.3em}&+\lambda^2/6\color{red}+\lambda^3/12+\lambda^4/36+\cdots&\color{black} = 1+\lambda+7\lambda^2/12+\cdots\end{array}
and use long division

\begin{array}{r}1-\lambda^2/12+\cdots\\ 1+\lambda+7\lambda^2/12+\cdots \,{\overline{\smash{\big)}\,1+\lambda+\phantom{9}\lambda^2/2\phantom{9} +\cdots}}\\ \underline{1+\lambda+7\lambda^2/12+\cdots }\\0-\lambda^2/12+\cdots\\ \underline{-\lambda^2/12+\cdots}\\0+\cdots \end{array}
to show

\mathcal{S} = \frac{1}{\lambda^2} \left(1-\frac{\lambda^2}{12}+\mathcal{O}[\lambda^4]\right) =\frac{1}{\lambda^2}-\frac{1}{12}+\mathcal{O}[\lambda^2].
For finite \lambda, the difference

\mathcal{S}-\mathcal{I}=\sum_{n=0}^\infty n\, e^{-\lambda n}-\int_{0}^\infty \hspace{-0.8em} dx\, x\,e^{-\lambda x} = \frac{1}{\lambda^2}-\frac{1}{12}+\mathcal{O}[\lambda^2]-\frac{1}{\lambda^2}= -\frac{1}{12}+\mathcal{O}[\lambda^2],
and for vanishing \lambda

\color{red}\lim_{\lambda \rightarrow 0}\color{black}\left( \sum_{n=0}^\infty n\, \color{red}e^{-\lambda n}\color{black}-\int_{0}^\infty \hspace{-0.8em} dx\, x\, \color{red}e^{-\lambda x} \color{black}\right) = -\frac{1}{12}.
Thus, removing nonphysical infinity in a controlled way exposes a finite component

\infty - \infty \stackrel{I}{=} -\frac{1}{12}
of the divergent natural numbers sum. To celebrate this hidden “golden nugget”, write

1+2+3+\cdots \stackrel{R}{=} -\frac{1}{12},
where”R” denotes **regularized**, **renormalized**, and **remainder**. “R” also denotes **Ramanujan**, who discovered this **association** without any formal mathematical training, and **Riemann**, whose famous **zeta function** \zeta[s] provides an alternate path to it.

In brief, use the convergence factor to tame infinity by isolating the diverging, remaining, and vanishing parts of the natural numbers sum

\sum_{n=1}^\infty n=\color{red}\lim_{\lambda \rightarrow 0}\color{black}\sum_{n=0}^\infty n\color{red}\, e^{-\lambda n}\color{black}=\color{red}\underbrace{\ +\frac{1}{\lambda^2}\ \ }_\text{Diverge}\color{black}\underbrace{\ -\frac{1\vphantom{\lambda^2}}{12}\ \ }_\text{Remain}\color{red} \underbrace{\vphantom{-\frac{1\vphantom{\lambda^2}}{12}}+\mathcal{O}[\lambda^2]}_\text{Vanish}\color{black}\ \stackrel{R}{=}-\frac{1}{12}.
Discard the first, retain the second, and let go the third.