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How to prove that

$\displaystyle \left( \sum_{k=0}^{\infty }\frac{\left( -a\right) ^{k}y^{2k}}{k!}\right) \left( \sum_{k=0}^{\infty }\frac{a^{k}y^{2k+1}}{\left( 2k+1\right) k!}% \right) =\sum_{k=0}^{\infty }\frac{2^{2k}\left( -a\right) ^{k}k!}{\left( 2k+1\right) !}y^{2k+1}$.

The coefficient of $y^{2k+1}$ can be written as $\displaystyle \sum_{i=0}^{k}\frac{\left( -a\right) ^{k-i}y^{2\left( k-i\right) }}{\left( k-i\right) !}\frac{a^{i}y^{2i+1}}{\left( 2i+1\right) i!} $ thus, it is remaining to prove that $\displaystyle \sum_{i=0}^{k}\frac{\left( -1\right) ^{i}}{\left( k-i\right) !i!\left( 2i+1\right) }=\frac{2^{2k}k!}{\left( 2k+1\right) !} $

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    This is against the site policy to delete one's own question for no clear reason. Previous version restored.2012-08-28

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HINT Look at the ways an odd number can be written as a sum of an even number and another odd number i.e. $2k+1 = 0 + 2k+1 = 2 + 2k-1 = \cdots = 2k-2 + 3 = 2k +1$

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    Thanks for your hint. But it is just a comment not an answer. Can you edit it so that I can delete this question?2012-08-08