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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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In your first, I would make the dummy variables different. That way, if you combine the two sums on the left, you can keep track of them more easily. – Ross Millikan Jun 23 '12 at 21:11
This is against the site policy to delete one's own question for no clear reason. Previous version restored. – Did Aug 28 '12 at 22:58

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? – Xiangyu Meng Aug 8 '12 at 5:45

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