Confusion using approximations to solve limits

Question

I'm an algebraist and have always been amazed by the mileage analysts are able to get out of approximations, and in particular how they are able to reason formally about big and little o terms.

I was prompted to write this question when a read a comment in this MSE question from today:

The question asks why $\prod (1 + \frac{1}{2^n})$ converges, and a commenter says to use the approximation $\ln(1+x) \approx x$.

This is clearly useful, since

$$ \ln \left (\prod(1+\frac{1}{2^n}) \right) = \sum \ln \left (1+\frac{1}{2^n} \right ) \approx \sum \frac{1}{2^n} = 2 $$

To me, this gives a heuristic that the product should converge to $e^2$, but shouldn't count as a real proof. The approximation we used sweeps a bunch of error terms of the form $\frac{x^n}{n}$ under the rug, and it's not clear to me how we would even get a bound, since the error terms alternate in sign.

I'm intending for this problem, which I found myself unable to answer, to be treated as a case study. I'm more interested in the method and the tools than in this one product.

How does one make this rigorous? What happens to the $O((\frac{1}{2^n})^2)$ term in each position? How small do we need to be to use these approximations in general? And can we compute the exact sum using these tools, or can we only get convergence from them?

Sorry if this should be obvious, I guess I'm one of today's lucky 10,000.

Thanks in advance ^_^

Answer 1

It is (almost) perfectly rigourous, because the notion involved is not approximation, but asymptotic equivalence. It relies on a standard theorem which is is one of the most powerful tools to determine convergence or divergence of series:

Let $\sum_n a_n$, $\sum_n b_n$ two series with ultimately positive terms. If $a_n\sim_\infty b_n$, then $\sum_n a_n$ and $\sum_n b_n$ both converge or both diverge.

Of course, it can be generalised to series with ultimately negative terms.