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If $R$ is noetherian, then $R/m^k$ is an $R$-module of finite length, (Eisenbud, Thm 2.14, p74.) Hence there exists a $\lambda$ such that the image of $a_{\mu}$ in $R/m^k$ is equal to the image of $a_{\lambda}$ for all $\mu\ge \lambda$. Thus the image of $a_{\lambda}$ is equal to the image of the intersection of all $a_{\mu}$ $\mu\ge \lambda$, which is zero. Which means $a_{\lambda}$ is contained in $m^k$.

If $R$ is not noetherian then it is not true, e.g $R=k[[X_1,...]]$ modulo all the monomials of degree $2$, so that $m^2=0$, $a_i= (X_i, X_{i+1}, ....)$. Then none of the $a_i$ is contained in $m^2$.

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If $R$ is noetherian, then $R/m^k$ is an $R$-module of finite length, (Eisenbud, Thm 2.14, p74.) Hence there exists a $\lambda$ such that the image of $a_{\mu}$ in $R/m^k$ is equal to the image of $a_{\lambda}$ for all $\mu\ge \lambda$. Thus the image of $a_{\lambda}$ is equal to the image of the intersection of all $a_{\mu}$ $\mu\ge \lambda$, which is zero. Which means $a_{\lambda}$ is contained in $m^k$.

If $R$ is not noetherian then it is not true, e.g $R=k[[X_1,...]]$ modulo all the monomials of degree $2$, so that $m^2=0$, $a_i= (X_i, X_{i+1}, ....)$. Then none of the $a_i$ is contained in $m^2$.