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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

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    $\begingroup$ Hint: Fubini. (Unrelated: I noticed that you often call "My idea" a pragraph which does not describe an "idea", at most a trivial rewriting of the question leading nowhere nearer to a solution. In such cases, you might want to adopt another terminology?) $\endgroup$ – Did Jun 14 '14 at 8:38
  • $\begingroup$ Oh,@OmranKouba,Thank you very much! $\endgroup$ – math110 Jun 14 '14 at 15:37
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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 $$ \frac{\pi^3}{648}=\sum_{n=1}^\infty\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{4n}(2n+1)} $$

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