<s>[INST] Prove that for $f(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{2^n+1}$ 
Prove that for $f(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{2^n+1}$,
$$\lim_{z \to2} f(z) (2-z) = 2$$

My approach : 
I was thinking of applying Abel's limit theorem to compute the limit but since it is valid only for $z \to 1 $ I am totally clueless 
I also tried by this approach:
$$ \lim_{\epsilon \to 0 } f(2-\epsilon)(\epsilon)= \sum_{0}^{\infty} \frac{(2-\epsilon)^n}{2^n +1}(\epsilon)$$ 
but again i am not sure of how to proceed. 
 [/INST] Just note that for any $z \in (-2,2)$, because of absolute convergence in $[-\frac{2+z}{2},\frac{2+z}{2}]$, we have $$f(z)(2-z) = \sum\limits_{n=0}^{+\infty} \frac{2z^n}{2^n+1} - \sum\limits_{n=0}^{+\infty} \frac{z^{n+1}}{2^n+1} = 1 + \sum\limits_{n=0}^{+\infty} z^{n+1} \cdot \frac{2(2^n+1)-(2^{n+1}+1)}{(2^n+1)(2^{n+1}+1)} = 1 + \sum\limits_{n=0}^{+\infty} z^{n+1} \cdot \frac{1}{(2^n+1)(2^{n+1}+1)}$$ 
So, since this has radius $4$, it is continuous in $2$ (in $2^-$).
Thus $$\lim\limits_{z \to 2^-} f(z)(2-z) = 1 + \sum\limits_{n=0}^{+\infty} 2^{n+1} \cdot \frac{1}{(2^n+1)(2^{n+1}+1)} = 1 + 2\sum\limits_{n=0}^{+\infty} \frac{1}{(2^n+1)} - \frac{1}{(2^{n+1}+1)}$$
Therefore the result is $2$.
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