Question

Old problem confirm
\[\sum_{n=1}^\infty \dfrac{z^{n^2}}{n}=\sum_{k=1}^\infty a_kz^k\]
Then \(a_k =?? \)
Please, help

Answer 1

@SithsAndGiggles

Answer 2

If you're replacing \(n^2\) with \(k\), that means you have this pattern:
\[\begin{array}{c|cccc}
n&1&2&3&4&5&\cdots\\
\hline
k&1&4&9&16&25&\cdots
\end{array}\]
So while \(a_n\) is defined for all positive integers \(n\), \(a_k\) is only defined for perfect squares. In other words, you can transform \(a_n\) to
\[a_k=\begin{cases}a_{\sqrt n}&\text{ for }n\in\{1,4,9,16,25,\ldots\}\\0&\text{otherwise}\end{cases}\]