Complex Power series with factorials

Question

Find the radius of convergence of $$\displaystyle\sum_{n=0}^\infty z^{n!}$$ $$\displaystyle\sum_{n=1}^\infty {(-1)^nz^{n(n+1)}}/{n}$$ What is the behavior of the series for $z=1, -1,i$

Answer 1

$\sum\limits_{n=0}^\infty z^{n!}\leq\sum\limits_{n=0}^\infty z^n<\infty$ for $0\!<\!z\!<\!1$, but $\sum\limits_{n=0}^\infty 1^{n!}=\infty$, therefore the radius of convergence is $1$.

$\sum\limits_{n=1}^\infty\left|\frac{(-1)^n}nz^{n(n+1)}\right|\leq\sum\limits_{n=1}^{\infty}z^{n(n+1)}\leq\sum\limits_{n=1}^\infty z^n<\infty$ for $0\!<\!z\!<\!1$, but $\sum\limits_{n=1}^\infty\left|\frac{(-1)^n}n1^{n(n+1)}\right|=\sum\limits_{n=1}^\infty\frac1n=\infty$, so the radius of convergence is also $1$.

What was needed:

• The geometric series $\sum_{n=0}^\infty z^n$ is convergent for $|z|<1$.
• The harmonic series $\sum_{n=1}^\infty\frac1n$ is divergent.
• Absolute convergence implies regular convergence.
• Every power series has a circle of convergence inside which it's absolutely convergent and is divergent outside.