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The product converges normally: Pick $K=D(0,r)\subset D(0,1)$ and let $|\cdot|_K=\sup_K |\cdot|$ then

$$\sum_{n=1}^{\infty} |nz^n|_K\leq r|(\frac{1}{1-r}-1)'|=\frac{r}{(1-r)^2}<\infty$$

for each $0<r<1$. This implies that $f$ is holomorphic in $D(0,1)$

There is a typo sorry. It should say $\prod_{n=1}^{\infty}(1+nz)^n$.

---------- Post added at 14:28 ---------- Previous post was at 13:23 ----------

Even though my question was had a typo originally, how did you come up with this. I see that it will only converge if |z| < 1 so we need a geometric series but why do you have$$\sum_{n=1}^{\infty} |nz^n|_K\leq r|(\frac{1}{1-r}-1)'|=\frac{r}{(1-r)^2}<\infty$$

$$

r\left|\left(\frac{1}{1-r}-1\right)'\right| = \frac{r}{(1-r)^2}

$$

Why are we taking the derivative here as well?

I think there's something wrong: Take $z=\frac{1}{2}$ (the estimate works for any positive real number on the unit interval) then for $n>2$ we have $|(1+nz)|^n \geq 1+(nz)^n>2$ which cannot happen if the product converges since individual terms tend to $1$ in that case.

There is a typo sorry. It should say $\prod_{n=1}^{\infty}(1+nz)^n$.

As for your other question, just notice that $\sum_{n=0}^\infty nz^{n-1}=\left( \frac{1}{1-z}\right)'$ (taking the derivative term by term) and mind the indices.

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I think there's something wrong: Take $z=\frac{1}{2}$ (the estimate works for any positive real number on the unit interval) then for $n>2$ we have $|(1+nz)|^n \geq 1+(nz)^n>2$ which cannot happen if the product converges since individual terms tend to $1$ in that case.

As for your other question, just notice that $\sum_{n=0}^\infty nz^{n-1}=\left( \frac{1}{1-z}\right)'$ (taking the derivative term by term) and mind the indices.

The -1 is what I am unsure about. $\left|\left(\dfrac{1}{1-r}-1\right)\right|$. Where did it come from?

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Ok so I have an idea. $1 - nz^n\Rightarrow z = \sqrt[n]{\dfrac{1}{n}}$. So how do I show that each point on the unit circle is an accumulation point now?How can I show that each point on the unit circle is an accumulation?

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Fix $R = 1$. Let $N$ be such that $|z_N|\leq 2 < |z_{N + 1}|$.

Then for $|z| \leq 1$ and $n > N$ we have $\left|z/z_n\right| < 1/2$ and hence

$$

\left|\log\left[\prod\limits_{n = 1}^{\infty}(1 - nz^n)\right]\right| = \left|\sum_{n = 1}^{\infty}\log(1 - nz^n)\right|\leq 2\left(\frac{1}{z_n}\right)^{k_n = 2}.

$$

So the series

$$

\sum_{n = N + 1}^{\infty}\log(1 - nz^n)

$$

converges absolutely and uniformly when $|z|\leq 1$.