Denote $x=y/N$, your equation in terms of $y$ rewrites as $N^{-1}\sum_{k=0}^{N} (a_k N^{1-k})y^{k}=0$, you need a small root of such a polynomial in $y$. Denote $a_kN^{1-k}=b_k$, then $b_1$ is asymptotically constant, $b_0$ tends to 0 (I understand you so, please use $o$ instead of $O$ if it is the case), other $b_i$ are bounded (and also $b_N$ is very small but we do not use this.) You may use Rouchet theorem for the circle $|y|=c$ for small $c$ and the functions $f(z)=b_1z$, $g(z)=\sum_{j\ne 1} b_j z^j$. The function $f$ has exactly one root inside the circle $|y|=c$ and $|f|>|g|$ on the circle (for any fixed $c>0$ this is so if $b_0$ is small enough). Thus you get a root of your polynomial $f+g$ smaller than $c$.

Denote $x=y/N$, your equation in terms of $y$ rewrites as $N^{-1}\sum_{k=0}^{N} (a_k N^{1-k})y^{k}=0$, you need a small root of such a polynomial in $y$. Denote $a_kN^{1-k}=b_k$, then $b_1$ is asymptotically constant, $b_0$ tends to 0 (I understand you so, please use $o$ instead of $O$ if it is the case), other $b_i$ are bounded (and also $b_N$ is very small but we do not use this.) You may use Rouché theorem for the circle $|y|=c$ for small $c$ and the functions $f(z)=b_1z$, $g(z)=\sum_{j\ne 1} b_j z^j$. The function $f$ has exactly one root inside the circle $|y|=c$ and $|f|>|g|$ on the circle (for any fixed $c>0$ this is so if $b_0$ is small enough). Thus you get a root of your polynomial $f+g$ smaller than $c$.

Denote $x=y/N$, your equation in terms of $y$ rewrites as $N^{-1}\sum_{k=0}^{N} (a_k N^{1-k})y^{k}=0$, you need a small root of such a polynomial in $y$. Denote $a_kN^{1-k}=b_k$, then $b_1$ is asymptotically constant, $b_0$ tends to 0 (I understand you so, please use $o$ instead of $O$ if it is the case), other $b_i$ are bounded (and also $b_N$ is very small but we do not use this.) You may use Rouchet theorem for the circle $|y|=c$ for small$c$ and functions $f(z)=b_1z$, $g(z)=\sum_{j\ne 1} b_j z^j$. The function $f$ has exactly one root inside the circle $|y|=c$ and $|f|>|g|$ on the circle (for any fixed $c>0$ this is so if $b_0$ is small enough). Thus you get a root of your polynomial smaller than $c$.

Denote $x=y/N$, your equation in terms of $y$ rewrites as $N^{-1}\sum_{k=0}^{N} (a_k N^{1-k})y^{k}=0$, you need a small root of such a polynomial in $y$. Denote $a_kN^{1-k}=b_k$, then $b_1$ is asymptotically constant, $b_0$ tends to 0 (I understand you so, please use $o$ instead of $O$ if it is the case), other $b_i$ are bounded (and also $b_N$ is very small but we do not use this.) You may use Rouchet theorem for the circle $|y|=c$ for small  $c$ and the functions $f(z)=b_1z$, $g(z)=\sum_{j\ne 1} b_j z^j$. The function $f$ has exactly one root inside the circle $|y|=c$ and $|f|>|g|$ on the circle (for any fixed $c>0$ this is so if $b_0$ is small enough). Thus you get a root of your polynomial $f+g$ smaller than $c$.