Today I was wondering if the sum of the reciprocals of the Fibonacci numbers converged, and with the help of wikipedia, I learned that they did. What I found interesting was that the infinitely long series of fractions was irrational.
I couldn't find any other similar sums of fractions that converged to an irrational number. Does anybody know any other ones?
EDIT: Sums of fractions with an obvious pattern that converge to an irrational number.
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots=\frac{\pi^2}{6}$$
$$1-\frac12+\frac13-\frac14+\frac15-\cdots=\ln(2)$$
There are many examples of this, because many of the irrational numbers we like to talk about are values of transcendental functions, which can often be written as power series, so for example:
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$
and this particular equation works no matter what value of $x$ we plug in. Thus, plugging in $x=1$:
$$1+1+\frac1{2!}+\frac1{3!}+\frac{1}{4!}+\cdots=e^1=e$$
If you plug in a different rational $x$, you'll get another irrational number: $e^x$, written as a sum of rational fractions. (To be a bit more precise, $e^x$ is irrational for any non-zero rational $x$.) That's just one function that produces such series representations, and that's not the only way to obtain sums like these.
It's worth mentioning that every irrational number can be written as a sum of rational fractions. For example:
$$3+\frac{1}{10}+\frac{4}{100}+\frac{1}{1000}+\frac{5}{10000} + \cdots = \pi$$
There might not be a nice obvious pattern to the numerators, but it still should count, as it seems to meet the requirement.
\*hey\* produces *hey* (it's the same with dollar signs, since they want to take you into math mode on this site)
– pjs36
Mar 16 '18 at 00:45
Basel Problem: $$\sum_{i=1}^\infty \frac{1}{n^2}= \frac{\pi^2}{6}$$
Apery constant: $$\sum_{i=1}^\infty \frac{1}{n^3}= \zeta(3)$$
$$\frac{1}{1} + \frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+\cdots = e,$$ where $e$ is Euler's number.
You forgot the fact $Q$ is dense in $\Re$, so for every irrational number $r$, there's a sequence $({x_n})$ in $Q$ that converges to $r$ with $x_0 = 0$ (in other words: that starts in 0). Take $y_n = x_n-x_{n-1}$ (so $y_n \in Q$), then you have $\sum_{k=1}^{n} y_k = \sum_{k=1}^{n} x_k - x_{k-1} = x_n - x_0 = x_n \to r$, when $n \to \infty$. Remember that here $r$ is arbitrary. Just in case, $x_n$ is a Cauchy sequence in $\Re$, so $y_n$ converges to zero.
