# 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.

• Are you sure that Moore-Osgood Theorem applies there? (disclaimer, I have not idea what it is about) – user Nov 24 '18 at 18:52
• Why is $f_N$ uniformly convergent? – gammatester Nov 24 '18 at 18:54
• If we let $x$ vary in $[0,1]$ It seems to me that we have pointwise convergence and then this convergence is also uniform BUT: I believe I see the error! The limit function is not continuous as it is $1$ everywhere while it is $1/2$ in $1$! – Darkwizie Nov 24 '18 at 19:09
• $f_N(x) = \frac{1}{1+x^{N+1}}$ does not converge uniformly on $[0,1]$ which you need to apply that theorem. For example you can see this as as uniform convergence implies continuity of the limit and here $f_N(x) \to 1$ if $x<1$ and to $f_N(1) \to \frac{1}{2}$. – Winther Nov 24 '18 at 19:21
• @Winther Yes indeed! Good very interesting! Thank you – Darkwizie Nov 24 '18 at 19:22

In order to use the Moore-Osgood Theorem, you must make sure that $$(f_n)_{n \geq 0}$$ converges uniformly toward $$f$$.
$$i.e. \sup\limits_{[0,1]}|f_n - f| \rightarrow 0$$.