Is this the correct way to find the value of $(-1)^{1/2}$?

Question

Is this the correct way to find the value of $(-1)^{1/2}$:

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I used the Taylor series and then substituted $x=-2$ and $k=2$:

Answer 1

The following is a general theorem in power series theory (look up "Cauchy-Hadamard theorem" and "radius of convergence")

Let $\sum_{k=0}^\infty a_kz^k$ be a power series. Then:

• for all $z\in\Bbb C$ such that $\lvert z\rvert\limsup\limits_{n\to\infty}\lvert a_n\rvert^{1/n}<1$ the series converges;

• for all $z\in\Bbb C$ such that $\lvert z\rvert\limsup\limits_{n\to\infty}\lvert a_n\rvert^{1/n}>1$ the series does not converge.

Now, the Taylor series at $c=0$ of $(1+z)^\alpha$ is indeed $\sum_{k=0}^\infty \frac{\prod_{h=0}^{k-1}(\alpha-h)}{k!}z^k$, which we usually shorten by setting $\binom\alpha k:=\frac1{k!}\prod_{h=0}^{k-1}(\alpha-h)$.

Fact is that for all $\alpha\in\Bbb R\setminus\Bbb N$, $$\limsup_{n\to\infty}\left\lvert\binom\alpha n\right\rvert^{1/n}=\lim_{n\to\infty}\exp\left(\frac1n\sum_{k=0}^{n-1}\ln\frac{\lvert \alpha-k\rvert}{k}\right)\stackrel{\text{Stolz-Cesàro}}=e^{\lim_{n\to\infty}\ln\frac{\lvert \alpha-n\rvert}n}=e^0=1$$

So, we can safely assert that, if you actually went on to evaluate the partial sums, you'd see a non-convergent sequence. In fact, the general terms of the series will become arbitrarily large in module themselves, as you may be starting to see by inspecting their pattern: it's $-\frac{(2n-3)!!}{n!}$ (where by "$!!$" I mean the double factorial). Since $\frac{(2n+1)!!}{n!}\ge 2^{n}$, we have the rough estimate $$\frac{(2n-3)!!}{n!}\ge \frac{2^{n-2}}{n(n-1)}\to\infty.$$