If $x>0$we have $(1+x^2)f'(x)+(1+x)f(x)=1$ and $g'(x)=f(x), f(0)=g(0)=0$Prove: If $x>0$ we have $(1+x^2)f'(x)+(1+x)f(x)=1$. And $g'(x)=f(x), f(0)=g(0)=0$ 
Prove that:$\displaystyle \frac14<\sum_{n=1}^{\infty}g(\frac1n)<1$
I tried solving the ODE,But it seems very complex.and I still have no idea about it.Could someone help me?
Thanks!
 A: Using integrating factor $m$ we first solve given ODE:
\begin{align*}
f'+\frac{1+x}{1+x^2}f&=\frac{1}{1+x^2}\\
m'&=m\frac{1+x}{1+x^2}\\
(\ln(m))'&=\frac{1+x}{1+x^2}\\
\ln(m)&=\int_0^x\frac{1+y}{1+y^2}dy=arctg(x)+\frac{1}{2}\int_{1}^{1+x^2}\frac{1}{z}dz=arctg(x)+\frac{1}{2}\ln(1+x^2)\\
m&=\exp(arctg(x))+\sqrt{1+x^2}\\
f&=\frac{1}{m}\int_0^x\frac{m}{1+y^2}dy=\frac{1}{m}\Big(\int_0^xarctg'(y)\exp(arctg(y))dy+\int_0^x\frac{1}{\sqrt{1+y^2}}dy\Big)=\frac{1}{m}\Big(\exp(arctg(x))-1+\ln(x+\sqrt{1+x^2})\Big)
\end{align*}
Thus
\begin{align*}
g(x)&=\int_0^xf(y)dy=\int_0^x\frac{\exp(arctg(y))-1+\ln(y+\sqrt{1+y^2})}{\exp(arctg(y))+\sqrt{1+y^2}}dy
\end{align*}
Using monotonicity of $f$ we get that
\begin{align*}
g(1)&\leq\int_0^1\frac{\exp(arctg(1))-1+\ln(1+\sqrt{1+1^2})}{\exp(arctg(1))+\sqrt{1+1^2}}dy=\frac{\exp(\pi/4)-1+\ln(1+\sqrt{2})}{\exp(\pi/4)+\sqrt{2}}<0,575
\end{align*}
A: Here's a start.
It's midnight,
so I am stopping
with what I have.
$g(x)
=g(0) + \int_0^x f(t) dt
=\int_0^x f(t) dt
$.
Therefore
$g(1/n)
=\int_0^{1/n} f(t) dt
$.
Since
$(1+x^2)f'(x)+(1+x)f(x)
=1
$,
$f(x)
=\frac1{1+x}-f'(x)\frac{1+x^2}{1+x}
$.
Since
$(\frac{1+x^2}{1+x})'
=1-\frac{2}{(1+x)^2}
$,
$h(x)
=\frac{1+x^2}{1+x}$
is decreasing for
$0 \le x 
< \sqrt{2}-1
$.
Therefore
$1 \ge h(x)
\ge h(\sqrt{2}-1)
=\frac{1+(2-2\sqrt{2}+1)}{\sqrt{2}}
=\frac{4-2\sqrt{2}}{\sqrt{2}}
=2\sqrt{2}-2
=c
\approx .818
$.
Therefore
$f'(x) \ge f'(x)h(x) \ge c f'(x)
$
so
$\int_0^{1/n} f'(x) \ge \int_0^{1/n} f'(x)h(x)
\ge c \int_0^{1/n} f'(x)
$
or
$f(1/n)
\ge \int_0^{1/n} f'(x)h(x)
\ge c f(1/n)
$.
Therefore
$\begin{array}\\
g(1/n)
&=\int_0^{1/n} f(t)dt \\
&=\int_0^{1/n}dt(\frac1{1+t}-f'(t)h(t))\\
&=\int_0^{1/n}\frac{dt}{1+t}-\int_0^{1/n}f'(t)h(t)dt\\
&=\ln(1+1/n)-\int_0^{1/n}f'(t)h(t)dt\\
\end{array}
$
so that
$cf(1/n)
\le \ln(1+1/n)-g(1/n)
\le f(1/n)
$.
So,
if we can get bounds for
$f(1/n)$,
we can sum them
to get what we want.
$f'(0) = 1$.
Differentiating
$(1+x^2)f'(x)+(1+x)f(x)
=1
$,
$0
=(1+x^2)f''+2x f' + f(x)+(1+x)f'
$,
or
$(1+x^2)f''
=-f-(1+3x)f'
$.
Therefore
$f''(0) = -1
$.
For small $x$,
$f(x)
\approx x-x^2/2
$
so
$f(1/n)
\approx \frac1{n}-\frac1{2n^2}
$.
I'm stopping here
since I need sleep.
