More mysterious fractions [closed]

Question

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?

Answer 1

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.