Let $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ be a power series with radius of convergence 1 and sum $f$. Let $S \in \mathbb { C }$. The purpose of this question is to prove that $$\left( \lim _ { \substack { x \rightarrow 1 ^ { - } \\ x \in \mathbb { R } } } f ( x ) = S \text { and } a _ { n } = o \left( \frac { 1 } { n } \right) \right) \Rightarrow \left( \sum _ { n \geqslant 0 } a _ { n } \text { converges and } \sum _ { n = 0 } ^ { + \infty } a _ { n } = S \right) . \quad \text{(Weak Tauberian)}$$
In the rest of this question we suppose that $\lim _ { \substack { x \rightarrow 1 ^ { - } \\ x \in \mathbb { R } } } f ( x ) = S$ and that $a _ { n } = o \left( \frac { 1 } { n } \right)$.
(a) Prove that for all $n \in \mathbb { N } ^ { * }$ and $x \in \left] 0,1 \right[$, we have $$\left| S _ { n } - f ( x ) \right| \leqslant ( 1 - x ) \sum _ { k = 1 } ^ { n } k \left| a _ { k } \right| + \frac { \sup _ { k > n } \left( k \left| a _ { k } \right| \right) } { n ( 1 - x ) }$$
(b) Deduce (Weak Tauberian) by specifying $x = x _ { n } = 1 - 1 / n$ for $n \in \mathbb { N } ^ { * }$.