Real Analysis: viewing infinity as the end

Question

Alternating harmonic series, $\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+ \ldots$ converges to $\log(2)$,

and the rearranged series, $\frac11-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}-\frac{1}{12} +\ldots$ converges to $\frac{\log(2)}{2}$.

Since the rearrangement is a bijective function, it contradicts with my intuition to have different value when summed.

Therefore, I tried to understand this situation, treating infinity as the end ( taking the sum on the interval $[0, ∞]$ instead of $[1,∞)$. )

Then, the original series becomes: $\frac11-\frac12+\frac13-\frac14+...+\frac1∞-\frac1∞$

and the rearranged series becomes: $\frac11-\frac12-\frac14+\frac13-\ldots+\frac{1}{(2∞/3)}-\frac{1}{(4∞/3)}$

Taking the difference of sums: $\frac{1}{(2∞/3)} + \ldots + \frac1∞ + \ldots + \frac{1}{(4∞/3)}$

This equals to $\int_0^1\frac{1}{(2/3 + 2x)}$ dx which is $\frac{\log(2)}{2}$.

I wonder if this way of thinking is allowed or if it reveals my lack of understanding of infinity.

Answer 1

The Riemann series theorem shows that the terms of a conditionally convergent series can be rearranged to converge to any value you like. This would be a good place to start.

In general, unless you really know what you're doing, you never want to manipulate $\infty$ as if it had some value. When people write $1+\frac12+\frac13+\cdots=\infty$, for example, this is really just a shorthand for a limit: $$\lim_{n \to \infty}1+\frac12+\frac13+\cdots+\frac1n=\infty,$$ and even then, the "equality" is a shorthand for indicating that the left-hand side grows without bound. With that in mind, it doesn't make sense to subtract $\infty$ from both sides.