Find the mistake in $\lim_{x\rightarrow 1^-} \frac{\sum_{n=0}^\infty x^n}{\sum_{n=0}^\infty x^n}=1 \Rightarrow 1=\frac{1}{2}$ It is obvious that we have:
$$\lim_{x\rightarrow 1^-} \frac{\sum_{n=0}^\infty x^n}{\sum_{n=0}^\infty x^n}=\lim_{x\rightarrow 1^-}1=1.$$
But let us now write this sum in two ways, let $a_n=x^n$ and $b_n=x^{2n}+x^{2n+1}$ we thus have $\sum_{n=0}^\infty x^n = \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty b_n$. We can write the above limit as:
$$
\lim_{x \rightarrow 1^-}\lim_{N\rightarrow \infty} \frac{\sum_{n=0}^N a_n}{\sum_{n=0}^{N} b_n} = \lim_{N\rightarrow \infty} \lim_{x \rightarrow 1^-} \frac{\sum_{n=0}^N a_n}{\sum_{n=0}^{N} b_n},
$$
where we can swap limits because of the Moore-Osgood Theorem. We now find for the right hand side:
$$
\lim_{N\rightarrow \infty} \lim_{x \rightarrow 1^-} \frac{\sum_{n=0}^N a_n}{\sum_{n=0}^{N} b_n}=\lim_{N\rightarrow \infty}  \frac{N+1}{2(N+1)}=\frac{1}{2}.
$$
This shows that $1=\frac{1}{2}$ which is clearly incorrect, but I do not see where the error occurs, I guess it is in the step where the Moore-Osgood Theorem is applied where we define $f_N(x)=\frac{\sum_{n=0}^N a_n}{\sum_{n=0}^{N} b_n}$. 
EDIT: I believe I have found the error, in order to apply the Moore-Osgood Theorem we need uniform convergence from $f_N(x)$ to $f(x)=\frac{\sum_{n=0}^\infty a_n}{\sum_{n=0}^{\infty} b_n}$ but this $f$ is not continuous, therefore we can not apply Dini's theorem to show that pointwise convergence implies uniform convegence.

In order to use the Moore-Osgood Theorem, you must make sure that $(f_n)_{n \geq 0}$ converges uniformly toward $f$. 
$i.e. \sup\limits_{[0,1]}|f_n - f| \rightarrow 0$.
This is not the case here.
