Inspired by Archit Boobna

To find minimum value of "z", we may use the minimum value of ${ a }_{ n }$ , as each ${ a }_{ n }$ decreases as z decreases, so minimum value of ${ a }_{ n }$ will occur at minimum value of z making it real.

(Why is this true? Differentiate any ${ a }_{ n }$ with respect to z, it will always be positive if real.)

Let ${ b }_{ n }={ a }_{ n+1 }$ . Now all the terms of these series are common. So for them to exist, they both need to exist together. Let's assume ${ b }_{ n }$ exists for ${ b }_{ n-1 }$ greater than a certain value k.

Since ${ b }_{ n-1 }$ is behaving in the same way for ${ b }_{ n }$ as ${ a }_{ n-1 }$ is doing for ${ a }_{ n }$ , min( ${ a }_{ n-1 }$ ) such that ${ a }_{ n }$ is real is also "k".

So, $\min _{ All\quad terms\quad are\quad real }{ { a }_{ n-1 } } =\min _{ All\quad terms\quad are\quad real }{ { b }_{ n-1 } } \\$ $\therefore \min _{ All\quad terms\quad are\quad real }{ { a }_{ n-1 } } =\min _{ All\quad terms\quad are\quad real }{ { a }_{ n } } \\$ $\therefore \min _{ All\quad terms\quad are\quad real }{ { a }_{ 1 } } =\min _{ All\quad terms\quad are\quad real }{ { a }_{ 2 } } \\$

So minima of terms (and consequently z) such that all terms are real will occur when ${ a }_{ 1 }={ a }_{ 2 }$

$\left( Here\quad "z"\quad represents\quad minimum\quad value\quad of\quad z \right) \\ \\ \ln { z } =\log _{ \ln { z } }{ z } \\ \therefore \log _{ e }{ z } =\log _{ \ln { z } }{ z } \\ \therefore e=\ln { z } \\ \boxed { \min { z } ={ e }^{ e } }$

Crap, I was just about to post a question just like this...

Now let the series be defined with the complex logarithm. Does there exist $z_\mathcal{C} = \min \lbrace z \mid z, \mathcal{L} \in \mathbb{R} \rbrace$

where $\lim_{n \to \infty}{a_n} = \mathcal{L}$ ?