Build a Blaschke product such as $B^*(1)=\lim_{r\to 1}B(r)=0$ Build a Blaschke product such as $B^*(1)=\lim_{r\to 1}B(r)=0$?
We have $$B(z)=\prod_{n=1}^{\infty}\dfrac{|z_n|}{z_n}\dfrac{z_n-z}{1-\bar{z_n}z}$$
I Know that $|B^*|=1$ p.p. on $\partial D$  (which $D=\{z: |z|<1\})$
Perhaps $z_n=1-\dfrac{1}{n^2}$ be suitable but how can I show it?
 A: Let $a_n=1-n^{-2}$. Hence, for $r>a_n$ 
\begin{eqnarray*}
|B(r)|&=&\prod_{n=1}^\infty\frac{r-a_n}{1-a_n r}\\
&\leq& \prod_{n=1}^{N-1}\frac{r-a_n}{1-a_n r} \text{ finite product}\\
&\leq& \prod_{n=1}^{N-1}\frac{r-a_n}{1-a_n } \text{ Bigger denominator}\\
\end{eqnarray*}
For $r<a_{N}$, we have also have
\begin{eqnarray*}\displaystyle
|B(r)| &\leq& \prod_{n=1}^{N-1}\frac{a_{N}-a_n}{1-a_n } \\
&\leq& \prod_{n=1}^{N-1} \left(1-\frac{n^2}{N^2}\right) \\
&\leq&e^{\ln \prod_{n=1}^{N-1} \left(1-\frac{n^2}{N^2}\right) }\\
&\leq&e^{\sum_{n=1}^{N-1}\ln  \left(1-\frac{n^2}{N^2}\right) }\\
&\leq&e^{\sum_{n=1}^{N-1}   \left( -\frac{n^2}{N^2}\right) }\,\, \, (\ln x\leq x-1)\\
\end{eqnarray*}
and hence the result.
A: Here's a result that works for any bounded holomorphic function.
Thm: Let $0<a_1< a_2 < \cdots \to 1,$ and assume
$$\frac{a_{n+1}-a_n}{1-a_{n+1}}\to 0.$$
Suppose $f\in H^\infty(\mathbb D)$ and $f(a_n)\to 0.$ Then $\lim_{r\to 1^-} f(r)=0.$
Proof: Let $|f|\le M$ in $\mathbb D.$ Note that if $z\in D(a_n, a_{n+1}-a_n),$ then $d(z,\partial D) < 1-a_{n+1}.$ So Cauchy's estimates show that for any such $z,$
$$|f'(z)| \le \frac{M}{1-a_{n+1}}.$$
Thus for $r\in (a_n,a_{n+1}),$
$$|f(r)-f(a_n)| = |\int_{a_n}^r f' |\le \frac{M}{1-a_{n+1}}(r-a_n) \le \frac{M}{1-a_{n+1}}(a_{n+1}-a_n) \to 0.$$
The theorem follows, as does your result for a Blashke product with zeros at $a_n = 1-1/n^2, n=1,2,\dots$
