Can a diverging series have a finite convergent value?

Question

This might be an odd question to ask, $$a - a^2 + a^3 - a^4 + ... = \frac{a}{1+a}$$

I came across this realisation while trying to understand a feedback loop of a buffer circuit of an op-amp.

The relation every op-amp follows is $V_{out} = A (V_+ - V_-)$ where $A \rightarrow \infty$ and by setting the $V_{out}$ and $V_-$ to be equal, we get a scenario where $V_{in} \frac{A}{A+1} = V_{out}$ or $V_{in} \approx V_{out}$

Trying to visualize how this circuit would converge on this input, I tried seeing how to $V_{out}$ changes as a sequence.

Suppose initially $V_{in}$ is 0, then $V_{out 0} = 0$. Now if $V_{in} = \delta V$, then $V_{out1} = A (V_{in} - V_{out0}) = A\delta V$

In the next step of the sequence, $V_{out2} = A (V_{in} - V_{out1}) = (A-A^2)\delta V$

In general, $$V_{outn} = A (V_{in} - V_{out(n-1)}) = \delta V (A - A^2 + A^3 - .... + (-1)^n A^n)$$

We know that this should converge to $\delta V \frac{A}{A+1}$.

Henceforth, this divergent behaviour somehow converges to finite value.

I was hoping to understand this behaviour better, can anyone guide me on this?

Answer 1

Short Version: Yes, diverging series CAN have a finite convergent assigned value. And it's very useful

Long Version:

Simple algebraic manipulation can verify your divergent series. Namely because

$$ 1 + a + a^2 + a^3 + ... = \frac{1}{1-a}$$

Then the alternating

$$ 1 - a + a^2 - a^3 + ... = \frac{1}{1+a} $$

A proof of the finite version of these can be found here and when $|a|<1$ then taking $n \rightarrow \infty$ in those formulas gives the formulas above. Now these identities can be verified to be true if $|a|<1$ but stating they are true when $|a|\ge 1$ is a bit more controversial.

Often divergent series do come up in physics a lot, and from what I understand, one can almost always use these questionable manipulations and they are usually experimentally verifiable (but you have to be careful! your experiment will only give 1 answer but if you aren't sophisticated with your mathematical approaches you might misleadingly believe there are multiple (almost all incorrect) renormalizations). A notable example is the Casimir Effect which rests on $1+2+3+ ... = -\frac{1}{12}$ and that family of divergent series.

On a separate note, I do believe that your series literally converges in the $p$-adics which is a different number system than the real numbers. It might be fruitful to reconsider your circuit problem in the p-adics but this is a bit advanced for a beginner and also would require a lot of work to import things like Maxwell's Equations into the p-adic realm. I do believe there are mathematical physicists working on such things.

Answer 2

The actual recurrence relation you're seeking in this analysis is: $$s(n)=V_{n+1}=\lim_{A\rightarrow\infty}A(V_n-V_{n-1})$$ (You can shift the subindices how you like, just leave the actual sequence number $s(n)$ untouched). Set the first input $V_{in}=V_1=k$, with $V_0=0$, where $k\in\mathbb{R}$ is any real value (idk if complex-valued voltages are a thing) to get: $$s(1)=\lim_{A\rightarrow\infty}A(k-0)=\infty$$ $$s(2)=\lim_{A\rightarrow\infty}A(\infty-k)=\infty$$ $$s(3)=\lim_{A\rightarrow\infty}A(\infty-\infty)$$ $$\vdots$$ You can easily see that for $n\geq3$ the sequence $s(n)$ is undefined. Also, we can algebraically check that: $$\begin{array}&\small s(1)=\lim_{A\rightarrow\infty}&kA\\\small s(2)=\lim_{A\rightarrow\infty}&k(A^2-A)\\\small s(3)=\lim_{A\rightarrow\infty}&k(A^3-2A^2)\\&\vdots\end{array}$$ This suggests that the sequence has a more complicated form than that of a simple geometric series, so i doubt that the sequence has the nice closed form you described as its limit. As for the question, divergent series never converge, but algebraic manipulations can make the sum have a limiting value. You can check an example here.