A holomorphic function with infinitely many zeros in the unit disc 
Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_1, z_2, z_3, \dotsc, z_n, \dotsc$ are its zeros ($\vert z_k \vert$ $\lt1$ ),then
  $$\sum_{k=1}^\infty (1-\vert z_k \vert) \lt \infty$$
  [Hint:Use Jensen's formula.]  

Since Jensen's formula can be used when $f$ vanishes nowhere on the circle $C_R$. I notice that there exist an increasing sequence $r_n$ for   $\lim_{n\to \infty} r_n = 1$, and $f$ vanishes nowhere on each $C_{r_n}$.
Suppose $f(0) \neq 0$, then use Jensen's formula on each circle $C_r$ and get
$$
    \sum_{k=1}^{n_r} \log \vert z_k \vert
  =   \log \vert f(0) \vert
    + n_r \cdot \log r
    - \frac{1}{2\pi}
      \int_{0}^{2\pi}
      \log \vert f(re^{i\theta}) \vert
      \,\mathrm{d}\theta,
$$
where $n_r$ denotes the numbers of zeros inside the disc $C_r$.
But I don't know how to estimate the limit of $n_r \log r$ as $r$ tends to $1$.

Of course this is a theorem instead of an exercise in many complex books, so we may as well add MSE to the list of places one can look it up...
Don't pull out the $r$ from $\log(|z_k|/r)$. Instead look at it this way: Define $$\log^+(t)=\begin{cases}\log(t),&(t>1),
\\0,&(0<t\le1).\end{cases}$$
Note that $$\sum_k\log^+(r/|z_k|)=-\sum_{|z_k|<r}\log(|z_k|/r).$$
So (assuming wlog that $f(0)\ne0$) Jensen implies that $$\lim_{r\to1}\sum_k\log^+(r/|z_k|)<\infty.$$
Applying the Monotone Convergence Theorem to that sum shows that
$$\sum_k\log(1/|z_k|)<\infty.$$
Or a more elementary version of the same argument: Say $|f|\le c$. Fix $N$. If $r$ is close enough to $1$ that $|z_k|<r$ for $k=1,\dots, N$ then Jensen shows that $$\sum_{k=1}^N\log(|z_k|/r)
\ge\sum_{|z_k|<r}\log(|z_k|/r)\ge\log|f(0)|-\log(c).$$Since $N$ is fixed we can let $r\to1$: $$\sum_{k=1}^N\log|z_k|\ge\log|f(0)|-\log(c).$$So $\sum_{k=1}^\infty\log|z_k|>-\infty$.
