I have a question considering the convergence of the complex series: $$\sum_{n=1}^{\infty} a_{n}z^n $$, where $ a_{n} = \begin{cases} k^k, & \mbox{if } n=k^2 \\ 0, & \mbox{otherwise } \end{cases}$.
How do we approach these situations? Should I plug $n=k^2$ and $a_{n}=k^k$ and apply the root or ratio test? Or should I work with the formal definition of the radius of convergence?
On
If you are aware of harmonic numbers and digamma functions, then
S1=∑n=0p1n+1=ψ(p+2)+γ=Hp+1 S2=∑n=0p1n+z=ψ(p+z+1)−ψ(z) Now, considering the asymptotics for large values of p S1−S2=(ψ(z)+γ)+1−zp+z2+z−22p2+O(1p3) which make ∑n=0∞(1n+1−1z+n)=ψ(z)+γ You will find here some special values of the digamma function.
Note that$$\sqrt[n]{a_n}=\begin{cases}(k^k)^{1/k^2}=\sqrt[k]k&\text{ if }n=k^2\text{ for some }k\\0&\text{ otherwise.}\end{cases}$$So, $\limsup_n\sqrt[n]{a_n}=1$ and therefore the radius of convergence of the series converges is $1$. Actually, it converges if and only if $\lvert z\rvert<1$.