7
$\begingroup$

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 :)

  • 0
    @Fabian it must be simple because this problem can be solved by a child with the age of 13.2012-12-04

2 Answers 2

13

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} $

5

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 $1/2$.

  • 0
    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.2012-12-04