Let $x$ be a real number such that $|x|<1.$ Which of the following is false? Let $x$ be a real number such that $|x|<1.$ Which of the following is false?


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*If $x\in \mathbb Q$, then $\sum_{m\ge 0}x^m \in \mathbb Q$

*If $\sum_{m\ge 0}x^m \in \mathbb Q$ then $x\in \mathbb Q$

*If $x\notin \mathbb Q$ then $\sum_{m\ge 0}mx^{m-1} \notin \mathbb Q$

*$\sum_{m\ge 0}x^m/m $ converges in $\mathbb R$
My attempt:-
$\sum_{m\ge 0}x^m =\frac{1}{1-x}$, If $x\in \mathbb Q$ then $\frac{1}{1-x}\in \mathbb Q$
Similarly, If $\frac{1}{1-x} \in \mathbb Q\implies 1-x \in \mathbb Q \implies$ $x\in \mathbb Q$
Also we know that $\frac{1}{(1-x)^2}=\sum_{m\ge 0}mx^{m-1} \in \mathbb Q$, If $x=1+\frac{1}{\sqrt 2}$
 So, (3) is the False statement. But In answer key it was given that (2) is the answer. I am confused.

Everything you have done is correct except that you cannot take $x=1+\frac 1 {\sqrt 2}$ since $|x|<1$. Take $x=1-\frac 1 {\sqrt 2}$ instead. 
