consider a sequence $(b_n)_{n\in\mathbb{N}}$ that tends to zero as $n\to\infty$ and is positive, as well as bounded from above by $1$. For $z\in [0,1)$ I try to relate the analytic function $g(z) = \sum_{i=0}^{\infty}z^i b_i$ to the expression $\sum_{i=0}^{\infty}(z^i-z^{i+1})b_{i+1}$.
One attempt I have made so far is that $$\sum_{i=0}^{\infty}(z^i-z^{i+1})b_{i+1}=\sum_{i=0}^{\infty}(z^i-z^{i+1})b_{i} - \sum_{i=0}^{\infty}(z^i-z^{i+1})(b_i - b_{i+1}) = g(z) - zg(z) - \sum_{i=0}^{\infty}(z^i-z^{i+1})(b_i - b_{i+1})$$ but then I end up with $\sum_{i=0}^{\infty}(z^i-z^{i+1})(b_i - b_{i+1})$, which again I do not know how to compare to $g(z)$.
I would appreciate any help and ideas!
It is $\sum\limits_{i=1}^{\infty} z^{i}b_{i+1}-\sum\limits_{i=1}^{\infty} z^{i+1}b_{i+1}$. Write the first term as $\frac 1 z \sum\limits_{i=1}^{\infty} z^{i+1}b_{i+1}$. This is also equal to $\frac 1 z \sum\limits_{i=2}^{\infty} z^{i}b_i =\frac 1 z (g(z)-b_0-b_1z)$. Can you take ti from here?