How Find the sum $\sum_{n=2}^{\infty}\left(\sum_{j=1}^{n}\frac{1}{(2j-1)^2} \binom{2n}{n}\frac{1}{2^{2n}(2n+1)}\right)$ 
Prove or disprove 
  $$I=\sum_{n=2}^{\infty}\left(1+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots+\dfrac{1}{(2n-1)^2}\right)
\binom{2n}{n}\dfrac{1}{2^{2n}(2n+1)}=\dfrac{\pi^3}{48}-\dfrac{1}{6}$$
My try: since
  $$\binom{2n}{n}\dfrac{1}{2^{2n}(2n+1)}=\dfrac{(2n)!}{(n!)^2\cdot 2^{2n}(2n+1)}=\dfrac{(2n-1)!!}{2^n\cdot n!(2n+1)}$$

so
$$I=\sum_{n=2}^{\infty}\left(\sum_{j=1}^{n}\dfrac{1}{(2j-1)^2}\cdot\dfrac{(2n-1)!!}{2^n\cdot n!(2n+1)}\right)$$
Then I can't continue.Maybe this sum have closed form,Thank you


Lemma. For $x\in[-1,1]$, we have
  $$
\frac{1}{6}(\arcsin x)^3=\sum_{n=1}^\infty a_nx^{2n+1}
$$
  with
  $$a_n=\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{2n}(2n+1)}.$$

Proof. First, we will suppose that $x \in(-1,1)$. Recall that,
$$\arcsin x= \sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n+1}}{2^{2n}(2n+1)}\tag{1}$$
Now, let $f(x)=\frac{1}{6}(\arcsin x)^3$. Clearly,
$$\sqrt{1-x^2}f'(x)=\frac{1}{2}(\arcsin x)^2,$$
hence
$$(1-x^2)f''(x)-x f'(x)= \arcsin x \tag{2}$$
Now, if $\sum_{n=0}^\infty a_nx^{2n+1}$ (with $a_0=0,a_1=1/6$,) is the power series expansion of $f$ then
replacing in $(2)$ we get
$$
\sum_{n=0}^\infty((2n+3)(2n+2)a_{n+1}-(2n+1)^2a_n)x^{2n+1}=\arcsin x 
$$
comparing with $(1)$ we see that
$$
(2n+3)(2n+2)a_{n+1}-(2n+1)^2a_n=\binom{2n}{n}\frac{1}{2^{2n}(2n+1)}
$$
This is equivalent to
$$
2^{2n}\frac{(2n+3)(2n+2)}{(2n+1)\binom{2n}{n}}a_{n+1}-\frac{(2n+1)2^{2n}}{\binom{2n}{n}}a_n=\frac{1}{(2n+1)^2}
$$
or
$$
2^{2n+2}\frac{(2n+3)}{\binom{2n+2}{n+1}}a_{n+1}-\frac{(2n+1)2^{2n}}{\binom{2n}{n}}a_n=\frac{1}{(2n+1)^2}
$$
adding these equalities for $n=0,1,\ldots,m-1$ we get
$$
2^{2m}\frac{(2m+1)}{\binom{2m}{m}}a_{m}= \sum_{n=0}^{m-1}\frac{1}{(2n+1)^2}
$$
which is the desired formula for $a_m$. Finally, noting that $a_n=\mathcal{O}(n^{-3/2})$, we conclude that the power series $\sum a_mx^{2m+1}$ does converge as 
$x$ tend to $+1$ or $-1$. So by Abel's theorem we conclude that this power series
converges to $f(x)$ also for $x=1$ and $x=-1$, and the lemma follows.$\qquad\square$
In particular, taking $x=1$ we obtain
$$
\frac{\pi^3}{48}=\sum_{n=1}^\infty\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{2n}(2n+1)}
$$
taking away the first term, we obtain $I=\frac{\pi^3}{48}-\frac{1}{6}$.
Also, taking $x=\frac{1}{2}$ we get
$$
\frac{\pi^3}{648}=\sum_{n=1}^\infty\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{4n}(2n+1)}
$$
