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

Question

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.

Answer 1

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 $$\sqrt{2}=\sum\limits_{n=0}^\infty\binom{2n}{n}\left(\frac{1}{8}\right)^n$$

Answer 2

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

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.

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