First we prove that $M\le 4.$ Indeed, if $M>4$ then by considering the sequence $a_{n}=2^{n-1}\;\forall \;n,$ and the constant $m=4<M$ we see that \[a_{1}+a_{2}+...+a_{n+1}=2^{n+1}-1<2^{n+1}=4a_{n}\;\;\forall \;n\ge 1\]a contradiction! Hence, $M\le 4.$ Next, we prove $M=4,$ if that wasn't the case, then for some $m<4$ and a sequence $(a_{n})_{n\ge 1}$ of positive reals one has \[a_{1}+a_{2}+...+a_{n}+a_{n+1}\le ma_{n}\;\forall \; n\ge 1\]Now define \[b_{n}=\frac{a_{n+1}}{a_{n}}\;\;\textrm{and}\; c_{n}=\frac{a_{1}+a_{2}+...+a_{n+1}}{a_{n}}\;\;\forall \;n\ge 1\]then we obviously have $\;\;c_{1}=1+b_{1},\;\;1<c_{n}\le m$ and $c_{n+1}=\frac{c_{n}}{b_{n}}+b_{n+1}$ for all $n\ge 1.$ Finally, consider the constant $C=\sqrt{m}(2-\sqrt{m})>0$ and a positive integer $N>\frac{m}{C}$, then apply AM-GM as follows: \[c_{N+1}+c_{N}+...+c_{2}+c_{1}=b_{N+1}+1+\sum_{n=1}^{N}(\frac{c_{n}}{b_{n}}+b_{n})\ge 1+b_{N+1}+2\sum_{n=1}^{N}\sqrt{c_{n}}\]\[\Rightarrow m\ge c_{N+1}>c_{N+1}-b_{N+1}-1\ge \sum_{n=1}^{N}(1-(\sqrt{c_{n}}-1)^{2})\ge N(1-(\sqrt{m}-1)^{2})=NC\]a contradiction! Thus, $M=4$.