If , $\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$ . For , $a_i\in\mathbb Z^+$ & $a_i
$Q.$ If , $$\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$$  For , $a_i\in\mathbb Z^+$ & $a_i<i$ . Find $a_2+a_3+......+a_7$ ?
MY APPROACH :
We have , $$\frac{5}{7}=\frac{3!-1}{3!+1}=\frac{3!+1-2}{3!+1}=1-\frac{2}{3!+1} \Rightarrow 1-\frac{2}{7}$$
Now I've : $$1-\frac{2}{7}=\underbrace{\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}}_{\Lambda}$$ $$\Lambda=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\left(\frac{1}{4!}-\frac{1}{5!}\right)+\left(\frac{1}{5!}-\frac{1}{6!}\right)+\left(\frac{1}{6!}-\frac{2}{7}\right)$$   OR ,$$\frac{a_2}{2!}+\frac{a_3}{3!}+ \frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac{1}{2!}+\frac{2}{3!}+ \frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!}-\frac{1433}{7!}$$
DOUBT : In my solution , all the terms are following the conditions from the Question except $a_7$ not satisfy $a_7\in\mathbb Z^+$ and $a_7\in\{1,2,3,4,5,6\}$ only by this condition $a_i<i$
$\Rightarrow$ Where am I wrong ?

Another way, starting from the other end.

*

*Multiply by $\,7!\,$ then $\,3600 = 7! \cdot \frac{5}{7} = 7 \cdot (\dots)+a_7\, \implies a_7 \equiv 3600 \equiv 2 \pmod{7}\,$, so $\bbox[border:1px solid black]{\,a_7=2\,}$.


*Move the known term to the left-hand side $\,\frac{5}{7}-\frac{2}{7!} = \frac{257}{360}\,$ and multiply by $\,6!\,$, then $\,514 = 6!\cdot \frac{257}{360} = 6 \cdot (\dots) + a_6 \implies a_6 \equiv 514 \equiv 4 \pmod{6}\,$, so $\bbox[border:1px solid black]{\,a_6 = 4\,}$.


*Move the known term to the left-hand side $\,\frac{257}{360}-\frac{4}{6!}=\frac{17}{24}\,$ and multiply by $\,5!\,$, then $\,85 = 5!\cdot \frac{17}{24} = 5 \cdot (\dots) + a_5 \implies a_5 \equiv 85 \equiv 0 \pmod{5}\,$, so $\bbox[border:1px solid black]{\,a_5 = 0\,}$.


*Nothing to move to the left-hand side now, multiply by $\,4!\,$, then $\,17 = 4!\cdot \frac{17}{24} = 4 \cdot (\dots) + a_4 \implies a_4 \equiv 17 \equiv 1 \pmod{4}\,$, so $\bbox[border:1px solid black]{\,a_4 = 1\,}$.


*Move the known term to the left-hand side $\,\frac{17}{24}-\frac{1}{4!}=\frac{2}{3}\,$ and multiply by $\,3!\,$, then $\,4 = 3!\cdot \frac{2}{3} = 3 \cdot (\dots) + a_3 \implies a_3 \equiv 4 \equiv 1 \pmod{3}\,$, so $\bbox[border:1px solid black]{\,a_3 = 1\,}$.


*Move the known term to the left-hand side $\,\frac{2}{3}-\frac{1}{3!}=\frac{1}{2}\,$, so $\bbox[border:1px solid black]{\,a_2 = 1\,}$.
