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 
 A: By general principles both $f$ and $g$ can be developed into a power series in terms of $x$ around the origin. Comparing coefficients one finds that
$$f(x)=x-{x^2\over2}+?x^3,\qquad g(x)={x^2\over2}-{x^3\over6}+?x^4\ .\tag{1}$$
For the task at hand we need quantitative estimates for $g$ valid in the interval $0\leq x\leq1$. In this regard we shall prove the following
Lemma. $\qquad\qquad{\displaystyle {x^2\over6}\leq g(x)\leq{x^2\over2}\qquad(0\leq x\leq1)\ .}$
Using this Lemma one immediately obtains
$${1\over4}<{\pi^2\over36}={1\over6}\sum_{n=1}^\infty{1\over n^2}\leq\ \sum_{n=1}^\infty g\left({1\over n}\right)\ \leq{1\over2}\sum_{n=1}^\infty{1\over n^2}={1\over2}{\pi^2\over6}\leq{5\over6}\ .$$
Proof of the Lemma: From $(1)$ it follows that $f(x)<x$ for small $x>0$. I claim that
$$f(x)<x\qquad(0<x<\infty)\ .\tag{2}$$
Assume that  there is an $x>0$ with $f(x)\geq x$. Then by continuity of $f$ there is a minimal such $x$, call it $\xi\ $, and we have $f(\xi)=\xi$. The differential equation for $f$ then implies that
$$f'(\xi)={1-(1+\xi)\xi\over 1+\xi^2}<1\ ,$$
whereas the line $y=x$ has slope $1$ at $(\xi,\xi)$. 
Together it would follow that $f(x)>x$ immediately to the left of $\xi$ – a contradiction.
From $(2)$ we already can conclude that
$$g(x)=\int_0^x f(t)\ dt\leq{x^2\over2}\ .$$
On the other hand, $(2)$ and the differential equation for $f$ imply
$$f'(x)\geq{1-(1+x)x\over 1+x^2}=1-x{1+2x\over1+x^2}\qquad(x\geq0)\ .\tag{3}$$
Since
$1+2x=2+x^2-(1-x)^2\leq2(1+x^2)$ it follows from $(3)$ that
$$f'(x)\geq 1-2x\qquad(x\geq0)\ .$$
This implies
$$f(x)=\int_0^x f'(t)\ dt\geq x-x^2\qquad(x\geq0)$$
and finally
$$g(x)=\int_0^x f(t)\ dt\geq{x^2\over2}-{x^3\over3}={x^2\over2}\left(1-{2x\over3}\right)\geq{x^2\over6}\qquad(0\leq x\leq1)\ .$$
A: Just some ideas !!!. No more.

It's is equivalent to show
$\displaystyle{%
\left\vert\sum_{n = 1}^{\infty}
{\rm g}\left(1 \over n\right) - {5 \over 8}\right\vert < {3 \over 8}}
$
$$
\sum_{n = 1}^{\infty}{\rm g}\left(1 \over n\right)
=
\sum_{n = 1}^{\infty}
{1 \over n}\,{{\rm g}\left(1/n\right) - {\rm g}\left(0\right) \over 1/n - 0}
=
\sum_{n = 1}^{\infty}{1 \over n}\,{\rm g}'\left(\xi_{n}\right)
=
\sum_{n = 1}^{\infty}{1 \over n}\,{\rm f}\left(\xi_{n}\right)
\,,
\qquad\qquad
0 < \xi_{n} < {1 \over n}
$$
$$
\sum_{n = 1}^{\infty}{\rm g}\left(1 \over n\right)
=
\sum_{n = 1}^{\infty}{1 \over n}\,\left\lbrack\xi_{n}\,
{{\rm f}\left(\xi_{n}\right) - {\rm f}\left(0\right) \over \xi_{n} - 0}\right\rbrack
=
\sum_{n = 1}^{\infty}{\xi_{n} \over n}\,{\rm f}'\left(\mu_{n}\right)\,,
\qquad\qquad
0 < \mu_{n} < \xi_{n} 
$$
$$
\sum_{n = 1}^{\infty}{\rm g}\left(1 \over n\right)
=
\sum_{n = 1}^{\infty}{\xi_{n} \over n}\,
{1 - \left(1 + \xi_{n}\right){\rm f}\left(\mu_{n}\right) \over 1 + \xi_{n}^{2}}\,,
\qquad\qquad
0 < \mu_{n} < \xi_{n} < {1 \over n}
$$
A: 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.
