Maybe
Chris is teaching me the science of forensic mathematics - How did I do Chris? LOL
$$\\a_n=(n+1)/(2n+3)\\\\
a_n-a_{n-1}=1/99\\\\
so\\\\
a_{n-1}=(n-1+1)/(2(n-1)+3)\\\\
a_{n-1}=n/(2n+1)\\\\
so\\\\
a_n-a_{n-1}=1/99\\\\
\frac{n+1}{2n+3}-\frac{n}{2n+1}=\frac{1}{99}\\\\
99(n+1)(2n+1)-99n(2n+3)=(2n+3)(2n+1)\\\\
etc$$
If you want me to continue, just ask. 