Sophie Germain and series

The bottom factors (via the Sophie Germain identity ) as $((n+1)^2+n^2)((n-1)^2+n^2),$ so the summand can be rewritten as $\frac{4n}{4n^4+1} = \frac1{(n-1)^2+n^2} - \frac1{n^2+(n+1)^2}.$ It should now be clear that the series telescopes , to $\frac1{(1-1)^2+1^2} = 1.$

Notice that, $4n^{4}+1\,=\,(2n^{2}+2n+1)(2n^{2}-2n+1)$ and $4n\,=\,(2n^{2}+2n+1)\,-\,(2n^{2}-2n+1)$ . Now, define $S\,=\, \sum_{n=1}^{r} \dfrac{4n}{4n^{4}+1}$ . From, aforementioned identities, it follow that, $S\,=\,\sum_{n=1}^{r}\dfrac{(2n^{2}+2n+1)\,-\,(2n^{2}-2n+1)}{(2n^{2}+2n+1)(2n^{2}-2n+1)}$ .

So, $S\,=\,\sum_{n=1}^{r}\left[\dfrac{1}{2n^{2}-2n+1}\,-\,\dfrac{1}{2n^{2}+2n+1}\right]\,=\,\sum_{n=1}^{r} \left [\dfrac{1}{2n^{2}-2n+1}\,-\,\dfrac{1}{2(n+1)^{2}-2(n+1)+1}\right]\,=\,\sum_{n=1}^{r}\dfrac{1}{2n^{2}-2n+1}\,-\,\sum_{n=2}^{r+1}\dfrac{1}{2n^{2}-2n+1}\,=\,1-\dfrac{1}{2(r+1)^{2}-2(r+1)+1}\,=\,1-\dfrac{1}{2r^{2}+2r+1}$ .

Now, required sum $=\,lim_{r \to \infty} S\,=\,\lim_{r \to \infty} \left(1\,-\,\dfrac{1}{2r^{2}+2r+1}\right)\,=\,1$ .

By using Sophie Germain we can solve it easily