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

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.

• Consider the binomial expansion of $\sqrt{1+x}$ and put $x=1$, say. Mar 31, 2021 at 15:33
• @Tavish isn't the binomial series only valid for $\lvert x\rvert <1$? I see your point though, thank you very much for the simple example. Mar 31, 2021 at 15:34
• Sure, you can take $x=0.5$ for example then. I really was referring to the limit $x\to 1^-$. Mar 31, 2021 at 16:16

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

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.

• This is the best answer I think.+1 Mar 31, 2021 at 16:05
• Thank you very much, this answer is great! Mar 31, 2021 at 17:53

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.

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 $$\phi=\frac{13}{2^3}+\sum_{n=0}^{\infty}\frac{(-1)^{n+1}C_{n+1}}{2^{4n+7}}.$$

• Wow...I love the ones involving $\phi$! Thank you! Mar 31, 2021 at 17:53
• For the first series involving $\phi$, do you mean to start the summation at $n=1$? Apr 10, 2021 at 22:37
• @A-LevelStudent Yes of course, I'm sorry for the typo. The series would otherwise not converge because we'd have $F_0=0$ in the denominator. Here is an approximation for the first $100$ terms of this series for the golden ration. Apr 10, 2021 at 22:50
• I thought so. I really do like that series, I even asked a question about it today :) math.stackexchange.com/questions/4097383/… Apr 10, 2021 at 22:51
• @A-LevelStudent Nice question! I guess one could prove it with the recurrence relation of the Fibonacci numbers. To apologize for my typo ;), here is another series for the golden ratio involving $\pi$ (which converges very fast): $$\phi = 2\sum_{n=0}^\infty\frac{(-1)^n\pi^{2n}}{25^n(2n)!}$$ Apr 10, 2021 at 23:03

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

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 $$x=1$$ $$\frac{1}{\sqrt3} = \frac{1}{2} + \frac{1}{16} + \frac{3}{156}+\frac{5}{2048} + \dots$$

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.

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