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 ?

You did great!
We can absorb the negative sign by using the other terms.
\begin{align}
&\frac1{2!}+\frac2{3!} + \frac3{4!} + \frac4{5!} + \frac5{6!} - \frac{1433}{7!} \\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} + \frac5{6!} \color{blue}{- \frac{1435}{7!} + \frac2{7!}}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} + \frac5{6!}\color{blue}{ - \frac{205}{6!}} + \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} \color{blue}{-\frac{200}{6!}} + \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} \color{blue}{-\frac{204}{6!} + \frac{4}{6!}}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} + \frac4{5!} \color{blue}{-\frac{34}{5!}} + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!}  \color{blue}{-\frac{30}{5!}} + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} + \frac3{4!} \color{blue}{ -\frac{6}{4!} }+ \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} \color{blue}{- \frac3{4!} }  + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} \color{blue}{- \frac{4}{4!} + \frac1{4!} }  + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} + \frac2{3!} \color{blue}{- \frac{1}{3!}} + \frac1{4!}   + \frac{4}{6!}+ \frac2{7!}\\
&=\frac1{2!} +  \color{blue}{\frac{1}{3!} }+ \frac1{4!}   + \frac{4}{6!}+ \frac2{7!}\\
\end{align}
I noticed that $1433=1435-2$ and $1435$ is a multiple of $7$, $204$ is a multiple of $6$ and so on.
