Sum of series or What?

We can write the sum as follows $\frac{4}{4+1}+\frac{8}{64+1}+\frac{12}{324+1}+\frac{14}{1024+1}\dots=\sum_{n=1}^{\infty}{\frac{4n}{4n^4+1}},$ using the Sophie-Germain equation $x^4+4y^4=((x+y)^2+y^2)((x-y)^2+y^2), x=1, y=n$ we obtain a telescopic serie $=\sum_{n=1}^{\infty}{\frac{4n}{((1-n)^2+n^2)((1+n)^2+n^2)}}$ $=\sum_{n=1}^{\infty}{\frac{1}{((1-n)^2+n^2)}-\frac{1}{((1+n)^2+n^2)}}$ $=1 -\lim_{n\rightarrow \infty}{\frac{1}{(1+n)^2+n^2}}=\boxed{1}$

Split 4 as 5-1....8 as 13-5....12 as 25-13...(1,5)....(5,13)....(13,25)...are factors of the denominator....so we end up forming a telescoping series upto infinity with the first term of 1....which is also the answer!!! Cheers!!

$\displaystyle \frac 45 + \frac 8 { 65 } + \frac { 12 } { 325 } + \frac { 16 } { 1025 } + \cdots = \boxed { 1 }$