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Any hint about this expression :

$$\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot3 \cdot5 \cdot7}+\ldots+\frac{n}{1 \cdot 3 \cdot 5 \cdot7 \cdot \ldots \cdot (2n+1)}$$

Thanks :)

there must be a trick :)

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If you believe that there must be a trick, I guess you expect a simple result? – Fabian Dec 4 '12 at 23:48
@Fabian it must be simple because this problem can be solved by a child with the age of 13. – Iuli Dec 4 '12 at 23:55
up vote 12 down vote accepted

Note that $$ \frac{2k}{1\cdot3\cdot5\cdots(2k+1)} =\frac1{1\cdot3\cdot5\cdots(2k-1)} -\frac1{1\cdot3\cdot5\cdots(2k+1)}\tag{1} $$ Summing up both sides, where the right hand side telescopes, yields $$ \sum_{k=1}^n\frac{2k}{1\cdot3\cdot5\cdots(2k+1)} =1-\frac1{1\cdot3\cdot5\cdots(2n+1)}\tag{2} $$ Letting $n\to\infty$ and dividing by $2$ yields $$ \sum_{k=1}^\infty\frac{k}{1\cdot3\cdot5\cdots(2k+1)}=\frac12\tag{3} $$

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Your $a_n = \dfrac{n}{1 \cdot 3 \cdots (2n-1) \cdot (2n+1)} = \dfrac{n 2^n n!}{(2n+1)!}$.

Claim: $$\sum_{n=1}^m \dfrac{n 2^n n!}{(2n+1)!} = \dfrac12 - \dfrac{2^{m-1} m!}{(2m+1)!}$$

The claim can be easily shown by induction.

Hence, the sum converges to $1/2$.

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How do you get your claim in the first place? – Patrick Li Dec 4 '12 at 23:54
I figured it out how to get the claim. $a_n = \frac{2^{n-2}(n-1)!}{(2n-1)!} - \frac{2^{n-1}n!}{(2n+1)!}$. If you know the final answer, then it's easier to go backwards and write $a_n$ in the above form. – Patrick Li Dec 4 '12 at 23:59

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