# More mysterious fractions [closed]

Let's have the following sums of fractions.

$$\frac{36020}{7203}+ \frac{36025}{7204}+ \frac{36030}{7205}+ \frac{36035}{7206}+ \frac{36040}{7207}+ \frac{36045}{7208}=F$$

$$\frac{15363}{5120}+ \frac{15366}{5121}+ \frac{15369}{5122}+ \frac{15372}{5123}+ \frac{15375}{5124}+ \frac{15378}{5125}=K$$

The sums of these fractions are equal to the the irrational numbers $$F,K$$ up to the tenth decimal.

Can you find the little trick required to express these irrationals $$F,K$$ in a closed form?

• I don't know what's the intended solution (perhaps some form of Euler–Maclaurin formula), but it looks to me that this question is an example of what doesn't meet the guideline. There are infinitely many irrational numbers in the specified intervals, and the term "closed form" isn't even meaningful without context. – WhatsUp Nov 28 '20 at 20:30
• @ WhatsUp. My first question on mysterious fractions is still posted. Go to that posting and your question will be answered – Vassilis Parassidis Nov 28 '20 at 20:37
• I would say the previous one is already in the gray area. But there is difference: the previous one asks for an explanation, and this one asks for two numbers which can have infinitely many choices. I can machine-generate thousands of answers to your question per second. – WhatsUp Nov 28 '20 at 20:40
• It's not obvious to me what this one really adds to the other one that was already answered. But I'm hoping to at least provide something more in the answer, namely a decent error estimate. (The error is smaller than the most naive estimate would suggest, unless I've messed up my most naive estimate.) – Gareth McCaughan Nov 28 '20 at 20:57
• (Actually, I think I have messed up my most naive estimate.) – Gareth McCaughan Nov 28 '20 at 21:00

Let's deal with the general case that covers both Vassilis's previous question along similar lines and this one, and consider $$\sum_{i=0}^{k-1}\frac{n(a+i+1)}{a+i}$$. This is $$k$$ times the arithmetic mean of the fractions involved, which is very close to $$k$$ times their geometric mean, which is $$kn\left(\prod_{i=0}^{k-1}\frac{ a+i+1}{a+i}\right)^{1/k}$$, which telescopes to give $$kn\bigl(\frac{a+k}{a}\bigr)^{1/k}$$.
So in this case we have $$F=30\root6\of{\frac{7209}{7203}}$$ and $$K=18\root6\of{\frac{5126}{5120}}$$.
How good an approximation is this? Those fractions are approximately equally spaced. The largest minus the smallest is $$\frac{n(a+1)}{a}-\frac{n(a+k)}{a+k-1}=\frac{k-1}{a(a+k-1)}\simeq\frac{nk}{m^2}$$ where $$m$$ is the "middle" denominator; and the value in the middle is close to $$n$$. So the relative error is about the same as you'd get by using $$\exp\frac{1}{2h}\int_{1-h}^{1+h}\log t\,dt$$ as an approximation to $$1=\frac1{2h}\int_{1-h}^{1+h}t\,dt$$ where $$h=\frac{k}{2m^2}$$. A bit of calculation with Taylor series shows that that error is about $$\frac16h^2$$ so our relative error should be on the order of $$\frac{k^2}{24m^4}$$. E.g., in the case of quantity $$F$$ this is about $$\frac{6^2}{24\cdot7200^4}\simeq1.7\times10^{-14}$$; the actual relative error is about $$2.1\times10^{-14}$$, so we've got the order of magnitude right.
Another way of looking at it, which may be more generally useful: if you have a bunch of quantities all close to one another, WLOG they're all close to 1; then AM minus GM equals $$\left(1+\frac1n\sum\varepsilon_i\right)-\left(\exp\frac1n\sum\log(1+\varepsilon_i)\right)$$ which to the first order that doesn't cancel out equals $$\frac12\frac1n\sum\varepsilon_i^2$$; that is, half the variance. And now we can finish the job as above by saying that the variance of the numbers in an AP is approximately that of a corresponding uniform distribution.