# [SOLVED]converges to a holomorphic function on D(0,1)

#### dwsmith

##### Well-known member
$f(z) = \prod\limits_{n = 1}^{\infty}(1 - nz^n)$

So I a trying to show that $f$ converges to a holomorphic function on the open unit disc $D(0,1)$.

For some reason, I am just not understanding this section on Weierstrass Products.

#### Jose27

##### New member
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 $D(0,1)$

• dwsmith

#### dwsmith

##### Well-known member
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 ----------

$$\sum_{n=1}^{\infty} |nz^n|_K\leq r|(\frac{1}{1-r}-1)'|=\frac{r}{(1-r)^2}<\infty$$
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
$$r\left|\left(\frac{1}{1-r}-1\right)'\right| = \frac{r}{(1-r)^2}$$
Why are we taking the derivative here as well?

#### Jose27

##### New member

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

#### dwsmith

##### Well-known member
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?

#### dwsmith

##### Well-known member
Could we start over with this problem? I don't understand how you have came up with what you have. There are some intermediate steps that are missing that aren't readily apparent to me.

Thanks.

#### dwsmith

##### Well-known member
How can I show that each point on the unit circle is an accumulation?

#### dwsmith

##### Well-known member
How can I show that each point on the unit circle is an accumulation?
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?

#### dwsmith

##### Well-known member
Is this correct?

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 $|z|\leq 1$.