# Real Analysis: viewing infinity as the end

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.

• Welcome to MSE. Please have a look at this helpful MathJax reference and guide to learn how to typeset mathematical expressions. – Théophile Apr 12 '19 at 21:11
• If you want to work with infinity, use the Projective Real line, or append the one point $\infty$ to $\mathbb{R}_+.$ You cannot subtract infinities, though in the right contexts (e.g. measure theory) you might define $0 \cdot \infty = 0$ to make arguments simpler. In general, you either need to carefully define your number system or instead work with limiting behavior. You are being too cavalier and making amateur mistakes; this is not unexpected from someone new to working with extended reals, but err on the side of caution for now. – Brevan Ellefsen Apr 12 '19 at 21:14

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.