Let $f(z) = \sum_{n=0}^{\infty} a_n z^n$ be an analytic function on $|z|\leq 1$ such that $|f(z) | \leq 1$ I want to prove following problem.  

Let $f(z) = \sum_{n=0}^{\infty} a_n z^n$ be an analytic function on $B = \{z \in \Bbb C : | z| \leq 1\}$ such that $|f(z)| \leq 1$ for all $z \in B$. Then $$\left| \sum_{n=k}^{\infty} a_n z^n \right| \leq \frac{|z|^k}{1-|z|}.$$

My trial : 
Since $|z|\leq 1$, What I know is R.H.S becomes
\begin{align}
\frac{|z|^k}{1-|z|} = |z|^k + |z|^{k+1} + \cdots 
\end{align}
and L.H.S are 
\begin{align}
\left| \sum_{n=k}^{\infty} a_n z^n\right| = |a_k| |z^k| + |a_{k+1}||z|^{k+1} + \cdots
\end{align}
so for me the problem becomes showing that all $|a_k| \leq 1$. 
From $|f(z)| \leq 1$
\begin{align}
|f(z)| = \sum_{k=0}^{\infty} |a_k| |z^k| \leq 1
\end{align}
and I got stuck. 
How one can prove that problem?
 A: Note that $a_n=\frac{1}{2\pi i}\oint_{\partial B}\frac{f(z)}{z^{n+1}}dz$, so
$$|a_n|\le \frac{1}{|2\pi i|}\oint_{\partial B}\frac{\big|f(z)\big|}{|z|^{n+1}}|dz|\le \frac{1}{2\pi}\oint_{\partial B}\frac{1}{1^{n+1}}|dz|=\frac{2\pi}{2\pi}=1.$$
Hence
$$\left|\sum_{n=k}^\infty a_nz^n\right|\le \sum_{n=k}^\infty|a_n||z|^n\le \sum_{n=k}^\infty |z|^n=\frac{|z|^k}{1-|z|}$$
for complex numbers $z$ s.t. $|z|<1$.
Here is a way to achieve a stronger inequality.  Let $f_n(z)=\sum_{m=n}^\infty a_mz^m$.  Then 
$$\frac{f_n(z)}{z^{r+1}}=\frac{f(z)}{z^{r+1}}-\sum_{m=0}^{n-1}\frac{a_m}{z^{r+1-m}}.$$
Therefore for $r\ge n$,
$$a_r=\frac{1}{2\pi i}\oint_{\partial B}\frac{f_n(w)}{w^{r+1}}dw=\frac{1}{2\pi i}\oint_{\partial B}\frac{f(w)}{w^{r+1}}dw.$$
That is
$$f_n(z)=\sum_{m=n}^\infty a_m z^m = \frac{1}{2\pi i}\oint_{\partial B}\sum_{m=n}^\infty\frac{f(w)}{w^{m+1}}z^mdw=\frac{1}{2\pi i}\oint_{\partial B}\frac{f(w)}{w}\frac{(z/w)^n}{1-(z/w)}dw$$
That is,
$$f_n(z)=\frac{z^n}{2\pi i}\oint_{\partial B}\frac{f(w)}{w^n(w-z)}dw,$$
so that
$$\big|f_n(z)\big|\leq \frac{|z|^n}{|2\pi i|}\oint_{\partial B}\frac{\big|f(w)\big|}{|w|^n|w-z|}|dw|=\frac{|z|^n}{2\pi}\oint_{\partial B}\frac{1}{|w-z|}|dw|.$$
Now
\begin{align}\oint_{\partial B}\frac{1}{|w-z|}|dw|&=\int_{0}^{2\pi}\frac{1}{\big|e^{i\theta}-|z|\big|}d\theta\\&=\int_0^{2\pi}\frac{1}{\sqrt{\big(\cos\theta-|z|\big)^2+\sin^2\theta}}d\theta
\\&=\int_0^{2\pi}\frac{1}{\sqrt{1+|z|^2-2|z|\cos\theta}}d\theta
\\&=\int_{0}^{2\pi}\frac{1}{\sqrt{\big(1+|z|\big)^2-4|z|\cos^2\frac{\theta}{2}}}d\theta
\\&=\frac{4}{1+|z|}\int_0^{\pi/2}\frac{1}{\sqrt{1-\left(\frac{2\sqrt{|z|}}{1+|z|}\right)^2\sin^2\varphi}}d\varphi\\&=\frac{4}{1+|z|}K\left(\left(\frac{2\sqrt{|z|}}{1+|z|}\right)^2\right),\end{align}
where $\varphi=\frac{\pi}{2}-\frac{\theta}{2}$ and
$$K(z)=\frac{\pi}{2}\sum_{r=0}^\infty\frac{1}{2^{4r}}\binom{2r}{r}^2z^r$$
is the complete elliptic integral of the first kind with parameter $\sqrt{z}$.  Therefore, we in fact have
$$\left|\sum_{m=n}^\infty a_m z^m\right|=\left|f_n(z)\right|\le \frac{2}{\pi}\left(\frac{|z|^n}{1+|z|}\right)K\left(\frac{4|z|}{\big(1+|z|\big)^2}\right)$$
for all $z$ s.t. $|z|<1$.  This plot shows that the bound above is better than the bound given in the original problem.
There is an even better bound for $\left|f_1(z)\right|$.  It can be proven using the Schwarz lemma that for $|z|\le 1$,
$$\left|f_1(z)\right|\le \frac{2|z|}{1+\sqrt{1-|z|^2}}.$$
To show this, suppose that $f(0)=v$.  If $|v|=1$, then $f$ must be constant by the maximum modulus principle, so $f_1=0$.  The inequality is true vacuously.  If $|v|<1$, then define
$$g(z)=\frac{f(z)-v}{1-\bar{v}f(z)}$$
for all $z\in B$.  Then $g:B\to B$ and $g(0)=0$.  By the Schwarz lemma, the function $h:B\to \Bbb C$ defined by $$h(z)=\left\{\begin{array}{ll}\frac{g(z)}{z}&\text{if}\; z\in B\setminus\{0\}\\
g'(0)&\text{if}\;z=0\end{array}\right.$$ is a holomorphic function from $B$ to $B$.  That is
$$\frac{f(z)-v}{1-\bar{v}f(z)}=zh(z)$$
or
$$f(z)=\frac{zh(z)+v}{1+\bar{v}zh(z)}.$$
Hence
$$\frac{f_1(z)}{z}=\frac{f(z)-v}{z}=\big(1-|v|^2\big)\frac{h(z)}{1+\bar{v}zh(z)}.$$
Therefore
$$\left|\frac{f_1(z)}{z}\right|\le\big(1-|v|^2\big)\frac{\left|h(z)\right|}{\left|1+\bar{v}zh(z)\right|}.$$
Because $\big|h(z)\big|\le 1$, we get
$$\left|\frac{f_1(z)}{z}\right|\le\big(1-|v|^2\big)\frac{1}{1-|\bar{v}z|}=\frac{1-|v|^2}{1-|v||z|}.$$
The maximum value of $\frac{1-|v|^2}{1-|v||z|}$ for a fixed $|z|$ is attained when $|v|=\frac{1-\sqrt{1-|z|^2}}{|z|}=\frac{|z|}{1+\sqrt{1-|z|^2}}$, which leads to
$$\left|\frac{f_1(z)}{z}\right|\le \frac{2\left(1-\sqrt{1-|z|^2}\right)}{|z|^2}=\frac{2}{1+\sqrt{1-|z|^2}}.$$
This is equivalent to the desired inequality.  Note that this bound is sharp.  For a given value $R$ s.t. $0\le R< 1$, there exists $z\in B$ s.t. $|z|=R$ for which $\big|f_1(z)\big|=\frac{2R}{1+\sqrt{1-R^2}}$ if and only if $$f(z)=\frac{\eta z+\frac{R}{1+\sqrt{1-R^2}}\upsilon}{1+\frac{R}{1+\sqrt{1-R^2}}\bar{\upsilon}\eta z}$$
for some $\eta,\upsilon\in \partial B$ (where the inequality becomes an equality when $z=-R\bar{\eta}\upsilon$). 
