Summation of a series with 2 different methods gives 2 different answers

Question

The objective here is to find the value of $S$, where $S$ is given by, $$S = 1-{1\over2}+{1\over3}-{1\over4}+...$$ I did this using two methods, but both the methods give different answers.

Method 1:

Let $$S' = 1+{1\over2}+{1\over3}+{1\over4}+...$$ Now,

$1+{1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+...$

$\;\;\;\;-1\;\;\;\;\;\;\;-{1\over2}\;\;\;\;\;\;\;-{1\over3}\;\;\;...$

$\frac{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}{}$

$1-{1\over2}+{1\over3}-{1\over4}+{1\over5}-{1\over6}+...$

So, $$S = S'-S' = 0$$

Method 2:

Using the formula for infinite Geometric Progression, we can say $$ 1 + x + x^2 + x^3 + ... = \frac{1}{1-x}$$ Integrating both sides with respect to $x$ from $0$ to $t$ we get, $$[x+{x^2\over2}+{x^3\over3}+{x^4\over4}+...]^t_0 = [-ln(1-x)]^t_0$$ $$t+{t^2\over2}+{t^3\over3}+{t^4\over4}+... = -ln(1-t)$$ Putting $t = -1$, $$-1+{1^2\over2}-{1^3\over3}+{1^4\over4}-... = -ln(2)$$ $$1-{1\over2}+{1\over3}-{1\over4}+... = ln(2)$$ So, $$S=ln(2)$$ After some searching I found that the 2nd method gives the correct answer. But why is the 1st method wrong?

Answer 1

The series you denote by $S'$ is the divergent harmonic series. You're implicitly using the theorem that the limit of the sum (or difference) of two sequences is the sum (or difference) of their limits. Since $S'$ has no limit, this theorem doesn't apply.

Answer 2

There is no single correct value. This conditionally convergent series can be rearranged and summed to any value, see the example in https://en.wikipedia.org/wiki/Riemann_series_theorem

Answer 3

Furthermore, Riemann's rearrangement theorem asserts that for any conditionally , but not absolutely, convergent series $u_n$, there is a rearrangement $u_{\sigma(n)}$ of its terms such that the re-arranged series converges to any prescribed number, or $+\infty$ or $-\infty$.

Answer 4

I want to add a (minor?) point to the other answers. You considered ∑aₙ and ∑(-1)ⁿaₙ in the case aₙ=n⁻¹. What happens when one replaces this with aₙ=nˢ?

Answer: For any s except s=‒1 your argument “would work”.

(Quotes are needed since I assume a particular way of regularization of divergent series for s≥‒1. However, I never saw any other way meaningfully used in the case of power-of-n series.)

Indeed, for s<‒1 your approach works due to the convergence of ∑aₙ. For s>‒1, both series may be regularized by Dirichlet method; the key observation is that Dirichlet regularization is preserved by dilution (which replaces a₁+a₂+… by 0+a₁+0+a₂+0+…). (Note that the alternating series is also Chesaro/Abel/Borel summable — to the same answer as by Dirichlet. The Chesaro summability is almost completely elementary.)

However, for s=‒1 the Dirichlet regularization gives ∞ (due to a pole of ζ(z) when z=1). So in your calculation you get ∞‒∞, and you see that it is not necessarily 0.

Essentially, what you demonstrated is that when one gets ∞ with an “advanced” summation method, one should be as careful as with the naive summation method.