Separate odd-index and even-index denominator terms

Both $1/[2(n+1)]$ and $1/[2(n+2)]$ are strictly positive, so $A_n=3/4$ minus two positive terms is less than $3/4$.

$E_n=\sum_{r=1}^n1/[2r(2r+2)]=\tfrac14\sum_{r=1}^n1/[r(r+1)]=\tfrac14[1-1/(n+1)]$.

For odd index $j=2r-1$, $j(j+2)=(2r-1)(2r+1)=4r^2-1$. The even-index terms are $1/[2r(2r+2)]$. Thus $\sum_{r=1}^n1/(4r^2-1)=A_{2n}-E_n$.

The result is $3/4-1/[2(2n+1)]-1/[2(2n+2)]-1/4+1/[4(n+1)]=\tfrac12[1-1/(2n+1)]=n/(2n+1)$. The last even denominator is $2n(2n+2)$, as printed in the rendered source; no factor of two is omitted.