Are there any known infinite series of rational terms that are just irrational (not transcendental)?

Question

I have probably encountered hundreds of infinite series where each term is rational. In each case (as far as I can remember), the value of the infinite series was either rational or transcendental.

For example, some simple cases include: $$\begin{align} \sum_{r=1}^{\infty}\frac{1}{2^r}&=1\\ \sum_{r=1}^{\infty}\frac{1}{r^2}&=\frac{\pi^2}{6}\\ \sum_{r=1}^{\infty}\frac{1}{r^2+1}&=\frac{1}{2}(\pi\coth\pi-1). \end{align}$$ I realize that it's definitely not known that it's true that infinite series of rational terms can only be rational or transcendental, as otherwise we wouldn't say that $\zeta(3)$ and other constants are irrational; we'd immediately be able to say that they are transcendental, so I'm not asking that. I'm asking if anyone knows of any infinite series of rational terms that is just irrational, not transcendental.

Thank you for your help.

Answer 1

This answer provides one infinite series for the number $\sqrt 2$, as well as two infinite series for the famous golden ratio $\phi$. $$\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!}{(1-2n)(n!)^24^n}=1+\frac{1}{2}-\frac{1}{8}+\frac{1}{16}-\frac{5}{126}+...=\sqrt 2.$$

This follows from the power series of $\sqrt{x+1}$. For more infinite series that are equal to $\sqrt 2$, look here.

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Series with Fibonacci numbers

Let $F_n$ denotes the $n$th Fibonacci number. Then $$\phi=1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_nF_{n+1}}.$$

Series with Catalan's number

Let $C_n$ denote the $n$th Catalan number. Then $$\phi=\frac{13}{2^3}+\sum_{n=0}^{\infty}\frac{(-1)^{n+1}C_{n+1}}{2^{4n+7}}.$$

Answer 2

The Taylor series expansion of $\sqrt{1+x}$ at $x=1$ is irrational, not transcendental, and each term of the series is rational.

Answer 3

Series $$ \frac{1}{(4-x)^{1/2}} = \frac{1}{2} + \frac{1}{16}x + \frac{3}{156}x^2+\frac{5}{2048}x^3 + \dots $$ has rational coefficinets and radius of convergence $4$. Plug in $x=1$ $$ \frac{1}{\sqrt3} = \frac{1}{2} + \frac{1}{16} + \frac{3}{156}+\frac{5}{2048} + \dots $$

Answer 4

$$\sqrt2 = 1 + \frac{4}{10} + \frac{1}{100} + \frac{4}{1000} + \frac{2}{10000} + \frac{1}{100000} + \frac{3}{1000000} + \cdots$$

In general, $$\alpha = \lfloor\alpha\rfloor + \sum_{n=1}^\infty \frac{\lfloor 10^n \alpha\rfloor - 10 \lfloor 10^{n-1}\alpha\rfloor}{10^n} $$ so every real number $\alpha$ -- be it rational, algebraic, or transcendental -- is the sum of some infinite series of rationals.

Answer 5

You can derive a infinite series of rational terms for any algebraic irrational number using the binomial theorem.

i.e.

$$\sqrt[r]{x}=\sum _{n=0}^{\infty } \frac{ \Gamma \left(n+\frac{1}{r}\right)}{n! \,\Gamma \left(\frac{1}{r}\right)}\left(1-\frac{1}{x}\right)^n \tag{1}$$

$$\frac{1}{\sqrt[r]{x}}=\sum _{n=0}^{\infty } \frac{ \Gamma \left(n-\frac{1}{r}\right)}{n! \,\Gamma \left(-\frac{1}{r}\right)}\left(1-\frac{1}{x}\right)^n\tag{2}$$

both valid for rational $x>1$ and $r>1$

see my previous questions here and here

Answer 6

Let $$a_n = \lfloor 10^nb\rfloor 10^{-n} $$

Then $a_n$ converges to $b$

Where $b$ is any irrational number .

From the sequence $a_n$ one can easily create a series (by subtracting of the previous term ).

So $\sum z_{n}$ goes to $b$ ,where $z_n=a_{n+1}-a_{n}$ and all $z_n$ are rational (since all $a_{n}$ are rational)

So taking $b$ to be an irrational number which is not transcendental answers the question.