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Analytic Continuation for the case : $\sum_{k=1}^{\infty} 1$


INTRODUCTION $\ \ \ \ \ $Find a convergence on the sum $1+1+1+1+...$ through analytic continuation of the series as a special case of $1+x+x^2+x^3+x^3$.


ATTEMPT $\ \ \ \ \ $Simple enough, right? People have evaluated crazier things. First, I establish the simplified convergence for the general equation, $1+x+x^2+x^3+x^3$. I establish the first term $a_0,$ as $1$ and the common ratio, $r$, as $x$. Then, simply, the convergence is $\frac{a_0}{1-r}$, or, in our case, $\frac{1}{1-x},$ keeping in mind that $|x| < 1$. Thusly, our convergence diameter is $(-1,1)$.
        Next, we develop the Taylor Series of our convergence formula at $1/2$. Why not? $1/2$ is as good a value as any (within our convergence radius), I would assume. So, I develop the series, $[(\frac{1}{2})^{-1}]-[(\frac{1}{2})^{-2}(x-\frac{1}{2})]+[(\frac{1}{2})^{-3}(x-\frac{1}{2})^2]...$ Recognizing this is a geometric series, I record the new $a_0=2$ and that the new $r=-2(x-\frac{1}{2})$.
        Evaluating the new convergence diameter by solving for $x$ in the new inequality, $|-2(x-\frac{1}{2})|<1$ yields the diameter $(0,1)$. At this point, I am extremely confused because, not only have I not sucessfully continued the domain to include the value I desire($1$), but I have also shrunken the convergence diameter. I have done the exact opposite than what the point of analytic continuation is.
        What have I done wrong? Is it that this case is inherently impossible to continue analytically? Have I made a computation error somewhere in my method? Was $1/2$ a badly chosen value to evaluate the Taylor Series from (if so, why, and what's a good one)?


REVISION $\ \ \ \ \ $I now clearly see that there will always be a pole at $1$ due to the convergence formula $\frac{1}{1-x}$. However, I was thinking. What if, instead of being a special case of $1+x+x^2+x^3+x^3...$, the sum was a special case of the series, $1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+...$? The new convergence formula would have $a_0=1$ and $r=\frac{1}{x}$, so $\frac{a_0}{1-r}=x$. There's no longer a pole at $1$! Does this make the series analytically continuable? I have tried inserting values such as $2$, which converges to $2$, (because $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...=2$) as well as with 3 and 4. The method checks out. I notice that I am now playing with a very dangerous fire, a fire whose name is the Riemann Zeta Function. However, is this now possible?

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It is indeed, inherently impossible because of the rigidity of meromorphic functions. The analytic continuation you obtained for $\sum_{i=0}^\infty x^i$ is $\frac{1}{1-x}$, a function defined on all of $\mathbb C $ except at $1$ and has a simple pole there. This means that you can find a sequence $x_n$ converging to $1$ that has unbounded image. By the identity principle this means that no meromorphic extension of $\sum_{i=0}^\infty x^i$ can ever be even continuous at $x=1$. So it will always be a pole.

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  • $\begingroup$ Thanks for clearing this up for me. I'm new to this and appreciate your help. $\endgroup$ – dsillman2000 Aug 27 '16 at 23:28
  • $\begingroup$ I revised my question. I haven't tried calculating it yet. Before I do, is it now continuable? $\endgroup$ – dsillman2000 Aug 27 '16 at 23:59
  • $\begingroup$ It still does not matter. The given function still agrees with $\frac{1}{1-x}$ on a set with a limit point (Take your favourite interval where your function converges) so any meromorphic extension to a larger set must agree with $\frac{1}{1-x}$. $\endgroup$ – Ravi Nov 28 '16 at 21:58

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