7. Let $AB$ and $CD$ be two perpendicular diameters of $\odot O$, $E$ be a point on segment $OB$, and $OE=\frac{1}{n} BE$. The extension of line segment $CE$ intersects $\odot O$ at point $F$, and line segment $AF$ intersects $DO$ at point $G$. Then $\frac{DG}{GC}=$ $\qquad$
7. $\frac{1}{n+1}$.

Let the radius of the circle be $n+1$. From the given information,
$$
O E=1, C E=\sqrt{1^{2}+(n+1)^{2}} \text {. }
$$

Connect $D F$. Then $\angle C F D=90^{\circ}$.
Thus, $\triangle E O C \backsim \triangle D F C \Rightarrow \frac{E C}{D C}=\frac{O C}{C F}$.
So, $C F=\frac{2(n+1)^{2}}{\sqrt{(n+1)^{2}+1}}$,
$$
E F=C F-C E=\frac{(n+1)^{2}-1}{\sqrt{(n+1)^{2}+1}} \text {. }
$$

Draw $F H \perp B O$, with the foot of the perpendicular at $H$. Then
$\triangle C O E \backsim \triangle F H E$
$$
\Rightarrow \frac{F H}{O C}=\frac{E H}{O E}=\frac{E F}{C E}=\frac{(n+1)^{2}-1}{(n+1)^{2}+1} \text {. }
$$

Thus, $F H=\frac{(n+1)^{2}-1}{(n+1)^{2}+1}(n+1)$,
$$
E H=\frac{(n+1)^{2}-1}{(n+1)^{2}+1} \text {. }
$$

Also, $\triangle A G O \backsim \triangle A F H$, so
$$
\begin{array}{l}
\frac{O G}{A O}=\frac{F H}{A H}=\frac{\frac{(n+1)^{2}-1}{(n+1)^{2}+1}(n+1)}{(n+1)+1+\frac{(n+1)^{2}-1}{(n+1)^{2}+1}} \\
=\frac{n}{n+2}, \\
O G=\frac{n(n+1)}{n+2} . \\
\text { Therefore, } \frac{D G}{G C}=\frac{n+1-\frac{n(n+1)}{n+2}}{n+1+\frac{n(n+1)}{n+2}}=\frac{1}{n+1} .
\end{array}
$$