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.

  • $\begingroup$ Are you sure that Moore-Osgood Theorem applies there? (disclaimer, I have not idea what it is about) $\endgroup$ – user Nov 24 '18 at 18:52
  • $\begingroup$ Why is $f_N$ uniformly convergent? $\endgroup$ – gammatester Nov 24 '18 at 18:54
  • $\begingroup$ 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 $1$! $\endgroup$ – Darkwizie Nov 24 '18 at 19:09
  • $\begingroup$ $f_N(x) = \frac{1}{1+x^{N+1}}$ 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 $f_N(1) \to \frac{1}{2}$. $\endgroup$ – Winther Nov 24 '18 at 19:21
  • $\begingroup$ @Winther Yes indeed! Good very interesting! Thank you $\endgroup$ – 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 $f$.

$i.e. \sup\limits_{[0,1]}|f_n - f| \rightarrow 0$.

This is not the case here.


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