<s>[INST] How prove this $\frac{1}{4}<\sum_{n=1}^{\infty}g\left(\frac{1}{n}\right)<1$ let  $x>0$, and such
$$(1+x^2)f'(x)+(1+x)f(x)=1,g'(x)=f(x),f(0)=g(0)=0$$
show that
$$\dfrac{1}{4}<\sum_{n=1}^{\infty}g\left(\dfrac{1}{n}\right)<1$$
my  idea: we can find 
$$f(x)=e^{-\int\dfrac{1+x}{1+x^2}dx}\left(\int\dfrac{1}{1+x^2}e^{\int\dfrac{1+x}{1+x^2}dx}dx+C\right)$$
since 
$$\int\dfrac{1+x}{1+x^2}dx=\arctan{x}+\dfrac{1}{2}\ln{(1+x^2)}$$
then following is very ugly,can someone have good methods? Thank you 
 [/INST] Notice that
$$(1+x^2)f'(x)+(1+x)f(x)=1\tag1$$
implies
$$ \left(e^{\arctan x}\sqrt{1+x^2}f(x)\right)'=\frac{e^{\arctan x}}{\sqrt{1+x^2}},\tag2$$
and $f(0)=0$. Thus, for $x>0$, it holds that
\begin{align*}
g'(x)&=f(x)\\&=\frac{1}{e^{\arctan x}\sqrt{1+x^2}}\int_0^x \frac{e^{\arctan t}}{\sqrt{1+t^2}}{\rm d}t\\
&\le \frac{1}{e^{\arctan x}}\int_0^x e^{\arctan x}{\rm d}t\\
&= x.\tag3
\end{align*}
 Therefore
$$g(x)=\int_0^x g'(t){\rm d}t\le \int_0^x t{\rm d}t= \frac{x^2}{2}.\tag4$$
On the other hand, for $x\geq 0$, it also holds that
\begin{align*}
f'(x)=\frac{1-(1+x)f(x)}{1+x^2}\ge \frac{1-(1+x)x}{1+x^2}\ge 1-2x,\tag5
\end{align*}
which implies 
$$ f(x)=\int_0^x f'(t){\rm d}t\ge \int_0^x (1-2t) {\rm d}t=x-x^2.\tag6$$
Thus
$$ g(x)=\int_0^x f(t){\rm d}t\ge \int_0^x (t-t^2){\rm d}t=\frac{x^2}{2}-\frac{x^3}{3}.\tag7$$
Consequently
$$ \frac{1}{4}\leq\frac{\pi^2}{36}=\sum_{n=1}^{\infty}\frac{1}{6n^2}\leq\sum_{n=1}^{\infty}\left(\frac{1}{2n^2}-\frac{1}{3n^3}\right) \leq\sum_{n=1}^{\infty}g\left(\frac{1}{n}\right)\leq \sum_{n=1}^{\infty}\frac{1}{2n^2}=\frac{\pi^2}{12}\le 1,\tag8$$
which is desired.
</s>