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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.