# Can I make an infinte sum using rational numbers that makes an irrational but not transcendental number?

I looked a lot on the internet for examples and I tried to do it myself, but I haven't seen any infinite sums of rational numbers that equal for example something like square root of 10 or cube root of 2.

The famous ones that I've seen end up adding to integers or a fraction of pi, but I haven't seen any irrational numbers that aren't transcendental.

Is there a method where I can pick an irrational number and then make an infinite sum using rational numbers where they are equal?

like the continued fraction for square root of 10 is 3.666666666... can I use that somehow to make an expression for the infinite sum using rational numbers? What about if it goes to a cube root?

I'd also really like it if someone could make my question a little clearer using the right math words and such, I really did my best to describe it.

• $\sqrt{2}=1+0.4+0.01+0.004+0.0002+...$ – BigbearZzz May 6 '16 at 17:29
• See the binomial series expansion. – Starfall May 6 '16 at 17:42
• @Starfall "See the binomial series expansion." What for? – Did May 6 '16 at 17:50

Suppose you have a decimal expansion for your irrational number. We'll take $\sqrt2=1.41421\dots$ for example.

Then we can write

$$\sqrt{2}=1+\frac{4}{10}+\frac{1}{100}+\frac{4}{1000}+\frac{2}{10000}+\frac{1}{100000}+\dots$$

Each term in our sequence is rational (and moreover it is clear how to apply this idea to get an infinite sum of rationals to converge to any real number).

Edit: and I now see @BigbearZzz made this exact remark in a comment

Edit 2: Alternatively from the identity $$\frac1{\sqrt{1-4x}}=\sum\limits_{n=0}^\infty\binom{2n}{n}x^n,$$ we can plug in $x=\frac{1}{8}$ and get $$\sqrt{2}=\sum\limits_{n=0}^\infty\binom{2n}{n}\left(\frac{1}{8}\right)^n$$

• I didn't do a very good job describing my question, I wanted to know if there was a way I could do something like the infinite sum of 1/n^2 which is equal to pi^2/6 . Like make a compact form, the one that you wrote I already knew. Is it not possible to make it compact for irrational numbers that aren't transcendental? – user337560 May 6 '16 at 17:41
• @user337560 It depends a little what you mean by a compact form - for example we could write $$\sqrt2=1+\sum\limits_{n=1}^\infty\frac{\left\lfloor10^n\sqrt2\right\rfloor}{10^n}-\frac{\left\lfloor10^{n-1}\sqrt2\right\rfloor}{10^{n-1}}$$ – Peter Woolfitt May 6 '16 at 17:47
• That would work except that there are irrational numbers in the compact form instead of only rational ones. Is it possible to make the compact form using only rational numbers? I did my best to convey it in the title question but I don't think I've done a very good job with the confusion I've caused. – user337560 May 6 '16 at 17:51
• @user337560 Each of the terms in the compact form given is rational (note the use of the floor function) . . . – Noah Schweber May 6 '16 at 17:56
• @user337560 Oh, I have got one. We have the identity $$\frac1{\sqrt{1-4x}}=\sum\limits_{n=0}^\infty\binom{2n}{n}x^n$$ so we can plug in $x=\frac{1}{8}$ and get $$\sqrt{2}=\sum\limits_{n=0}^\infty\binom{2n}{n}\left(\frac{1}{8}\right)^n$$ I'll add this one to the actual answer. – Peter Woolfitt May 6 '16 at 18:11

$\sqrt 2 = (1+1)^{1/2}$

by the binomial theorem:

$(1+a)^{1/2}$$= 1 + (1/2) 1^{-1/2}a-(1/8) 1^{-3/2}a^2+ (3/48) a^3\cdots$

coefficient of the $n^{th}$ term:

$c_0 = 1\\c_n = c_{n-1}\frac{(1/2-n)}{n}$

when $n\ge 3, c_n = (-1)^{n+1}\frac{1*3*5*7...(n-2)}{2*4*6*8\cdots n}$

More generally, can you find a series that converges to the alegraic (non-trancendental) number of your choosing?

If $q$ is an algebraic number, $p(q) = 0$ for some polynomial $p$.

And Newtons method will frequenty (but not always) generate a sequence that coverges to q.

$x_n = x_{n-1} - \dfrac {p(x_{n-1})}{p'(x_{n-1})}$