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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 ^_^

Bernard
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HallaSurvivor
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  • $\ln(1 + x)$ is not only approximately $x$ for small positive $x$, it is also smaller than said $x$. Can you solve it now? – Alexander Geldhof Nov 13 '19 at 23:38
  • @AlexanderGeldhof - That does solve this one problem (since now we have an upper bound which converges), but I'm still unclear on why these methods work in general. Why should the alternating sum of error terms wind up being negative, and how can I quickly see things like this in the future? – HallaSurvivor Nov 13 '19 at 23:40
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    It doesn't have to be negative for this kind of proof to hold. One possible definition for $f(x) \approx g(x)$ near a point $x_0$ (possibly $\infty$) is that there are $N, M \geq 0$, so that close enough near $x_0$, we have $N \cdot g(x) \leq f(x) \leq M \cdot g(x)$. Converge of the lower and upper series then implies convergence of the 'sandwiched' series. – Alexander Geldhof Nov 13 '19 at 23:44
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    Convergent alternating sums are the best case scenario: successive pairs of partial sums bound the convergent. – Eric Towers Nov 13 '19 at 23:45
  • @EricTowers Would you mind elaborating more? – HallaSurvivor Nov 13 '19 at 23:50
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    Recall that summands of a convergent alternating series monotonically decrease in magnitude to zero. (This is Leibniz's test.) Partial sums ending with a positive term form a decreasing sequence of upper bounds. The partial sums ending with a negative term form an increasing sequence of lower bounds. Consequently, from any truncation, the next two partial sums bound the error on both sides. – Eric Towers Nov 14 '19 at 00:01
  • Just knowing that the summands decrease in magnitude is (while obvious in hindsight) really helpful. That singlehandedly clears up a lot for me! – HallaSurvivor Nov 14 '19 at 00:11
  • Mathematica tells me that the product (starting at $n=0$) is approximately $4.768462$ which is considerably less than $e^2\approx 7.389056$. – Math1000 Nov 14 '19 at 00:34
  • @EricTowers Plenty of convergent alternating series do not monotonically decrease in magnitude to zero. – zhw. Nov 14 '19 at 06:24
  • @zhw. - Surely after finitely many terms this must be the case? I would love an example if you have one – HallaSurvivor Nov 14 '19 at 06:28
  • $1/3-1/2+1/3^2-1/2^2+1/3^3-1/2^3+\cdots$ – zhw. Nov 14 '19 at 06:37

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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.

Bernard
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