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Easeus Data Recovery Wizard v16.9 Keygen Serial Key keygenQ:

Prove$z_n$ converges to $z$ and $|z_n| \leq |z|$ for all $n$

I was doing this problem

Let $z\in\mathbb{C}$. Define $z_n = z^n + z^{2n} +… +z^{n^2}$.
Prove that $z_n$ converges to $z$ and that $|z_n| \leq |z|$ for all $n$.

Hint: Let $|z| = d$, and let $d_n = |z^n|$. Then $d = d_1$, and $d_n \leq d$, so $\limsup d_n \leq d$.
Then it says use the Comparison Test to show that $d_n$ approaches $d$.
I don’t really understand what they mean by ” $d_n \leq d$, so $\limsup d_n \leq d$ “. I tried to show $d_n \leq d$ but it seems that $d_n \geq d$, and I couldn’t show $d_n \leq d$ by just using the hint.
I know $|z_n| \leq |z|$ because $|z_n| = |z^n + z^{2n} +… +z^{n^2}| \leq |z|^n \leq |z|$
and $z_n$ converges to $z$ because $|z^n| \to 0$ so $z^n = o(1)$. But still I can’t use the hint.
Can anyone explain the hint to me?
Thank you.

A:

I assume you are told that $\lvert z^n\rvert\le\lvert z\rvert$ for every $n$.
If $d_n$ is the distance of $z_n$ to $0$, then
$$d_n\le\lvert z\rvert\lvert z^n\rvert\le d_n\tag{1}$$
and therefore
$$d_n\

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