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A number $a \in \mathbb{R}$ is transcendental if there isn't a polynomial $f(x) \in \mathbb{Q}[x]$ s.t. $f(a) = 0$.

I was thinking of generalizing this to infinite series. We will call a number $a \in \mathbb{R}$ series-trancendental if there isn't a power series $f(x) \in \mathbb{Q}[[x]]$ s.t. $f$ converges in $a$ and $f(a)=0$.

I was wondering if there is now a series-trancsendental number? (For example, $\pi$ isn't series-trancsendental because $\sin(\pi)=0$)

What if we expand the definition to allow $a \in \mathbb{C}$?

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    what doees [[.]] symbol mean?? – 2017-01-27
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    How would you prove something like that? How would you show that there is such a number? – 2017-01-27
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    @SubhashChandBhoria if $R$ is a ring, $R[[X]]$ denotes the ring of formal power series in variable $X$ over $R$. – 2017-01-27
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    Maybe there is some connection with periods? Pi is a period – 2017-01-27

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There are no series transcendental numbers. Given $a$ we can write a series with rational coefficients that has $f(a)=0$. Let $f(x)=\sum_{i=0}^\infty b_ix^i$ Choose $b_0 \in \Bbb Q$ such that $|b_0 -a| \lt1$ Now choose $b_1 \in \Bbb Q$ so that $|b_0+b_1a| \lt \frac 12$, then choose $b_2$ so that $|b_0+b_1a+b_2a^2| \lt \frac 14$ and so on. We have found a series such that $f(a)$ converges to $0$. Similarly if $a$ is complex, we can write $c=a\overline a$ and find a series in $x^2$ that takes $f(-c)$ to zero.

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    What if we restrict ourselves to coefficients represented by computable functions? Or even rational functions of the index? – 2017-01-27
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    Then there will be by a cardinality argument. There are only countably many computable functions, so there will be only countably many series rational numbers. These will be the same as the computable reals, as this is a computation. – 2017-01-27