We have:

1n2+(n+1)2=12n2+2n+1=12⎛⎜ ⎜ ⎜⎝1n2+n+12⎞⎟ ⎟ ⎟⎠<12(1n2+n)=12(1n−1n+1)

Apply, we have:

15+113+125+...+1n2+(n+1)2<12(1−12)+12(13−14)+12(14−15)+...+12(1n−1n+1)=12(1−1n+1)<12

We have:

\(\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{2n^2+2n+1}=\dfrac{1}{2}\left(\dfrac{1}{n^2+n+\dfrac{1}{2}}\right)< \dfrac{1}{2}\left(\dfrac{1}{n^2+n}\right)=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)

Apply, we have:

\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2}\left(1-\dfrac{1}{2}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{n+1}\right)< \dfrac{1}{2}\)