I need to prove the following
Let $b_n \geq 0$ for $n \in \mathbb{N}$. If for every $r \in (0,1)$ is $\sum_{n \geq1}b_n r^{n}\leq M < \infty$ then $\sum_{n\geq 1} b_n \leq M$
The book I read just says let $r \to 1^{-}$ but I want I rigorous step by step verification to convince myself. So let me share with you my try to give a proof for the above.
Let $\epsilon > 0$, for every $N \in \mathbb{N}$ there is a $r_0(N,\epsilon)=r_0$ such that for $r_0<r<1$ $$\sum_{n =1}^N b_n -\sum_{n= 1}^Nb_n r^{n}<\epsilon$$
witch implies that for every $N \in \mathbb{N}$, $$\sum_{n =1}^N b_n <\epsilon + \sum_{n\geq 1}b_n r^{n}=\epsilon+M$$ So, since $\sum_{n =1}^N b_n <\epsilon +M<\infty$ for every $N$, if we let $N \to \infty$ we get that $\sum_{n\geq 1} b_n \leq M+\epsilon$
Finally, $\epsilon>0$ was chosen at random so $$\sum_{n\geq 1} b_n \leq M $$
So, let me know if this is a valid proof. Is there a shorter path, but not vague as the statement let $r \to 1^{-}$ ? Thank you.
A better proof: For any $m$ we have $ \sum\limits_{n=1}^{m}b_nr^{n} \leq M$. This is true for all $r <1$. Let $r \to 1-$ to see that $ \sum\limits_{n=1}^{m}b_n \leq M$. Since this is now true for all $m$ we get $ \sum\limits_{n=1}^{\infty}b_n \leq M$.