Convergence of $(1+z)^n$ where $z,n\in\mathbb C$ and $|z|=1$

Question

While trying to express $\cos^n x(n\in\mathbb C)$ in terms of multiple angles, I tried to use the method of complex numbers by applying de-Moivre's theorem on $z=e^{ix}$: $$2^n\cos^nx=\left(z+\frac1{z}\right)^n$$ $$\implies2^n\cos^nx=z^{-n}(z^2+1)^n$$ Now, I got the radius of convergence of the series given by $(1+z^2)^n$ to be $1$ by the ratio test:

$$(1+z^2)^n=1+nz^2+\frac{n(n-1)}{2!}z^4+\frac{n(n-1)(n-2)}{3!}z^6+\cdots$$ $$\implies r=\lim_{t\to\infty}\left|\frac{t+1}{n-t}\right|=1$$

But $|z^2|=1$, so this is the boundary case, hence the series may be converging or diverging. So, how do I determine that for what values of $z$, this series will converge?

Please try to give answers in an easy-to-understand language from the perspective of a high-school student since as far as determining convergence or divergence of a series is concerned, I only know about the $n^{\text{th}}$ term test, the ratio test and determining the radius of convergence of a series by the ratio test.

Answer 1

Let $z\in\mathbb{C}$, $|z|=1$, and $\nu\in\mathbb{C}\setminus\mathbb{Z}_{\geqslant0}$. Then $\sum_{n=0}^\infty\binom{\nu}{n}z^n$

• converges absolutely if $\Re\nu>0$;
• converges if $\Re\nu>-1$ and $z\neq-1$;
• diverges otherwise.

If it converges, its sum is $(1+z)^\nu$.

The idea here is that $(-1)^n n^{\nu+1}\binom{\nu}{n}$ has a non-zero limit as $n\to\infty$, since $$ (-1)^n (n+1)^{\nu+1}\binom{\nu}{n}=\prod_{k=1}^n\left(1-\frac{\nu+1}k\right)\left(1+\frac1k\right)^{\nu+1} $$ does (the term under the product is $1+O(k^{-2})$ as $k\to\infty$).

(The limit is known as $1/\Gamma(-\nu)$, but we don't need this fact here.)

This yields the absolute convergence of $\sum_{n=0}^\infty\binom{\nu}{n}z^n$ when $\Re\nu>0$ (its term is $O(n^{-1-\Re\nu})$ as $n\to\infty$), and the divergence when $\Re\nu\leqslant-1$ (since the term doesn't tend to zero).

For $-1<\Re\nu\leqslant0$, $$ (1+z)\sum_{n=0}^N\binom{\nu}{n}z^n=1+\binom{\nu}{N}z^{N+1}+\sum_{n=1}^N\left[\binom{\nu}{n}+\binom{\nu}{n-1}\right]z^n $$ has a limit as $N\to\infty$, since $\binom{\nu}{N}=O(N^{-1-\Re\nu})$ and $\binom{\nu}{n}+\binom{\nu}{n-1}=\binom{\nu+1}{n}=O(n^{-2-\Re\nu})$ by the above. Thus, if $z\neq-1$, the series $\sum_{n=0}^\infty\binom{\nu}{n}z^n$ converges. At $z=-1$ it diverges, since otherwise, by Abel's theorem, its sum would be equal to $\lim_{r\to 1^-}(1-r)^\nu$, which doesn't exist.

The claim about the sum being equal to $(1+z)^\nu$ follows again from Abel's theorem.