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Find the sum of this series :

$\sum\limits_{\scriptstyle 1 \leqslant x \leqslant 2009 \atop {\scriptstyle x+y=2010 \atop \scriptstyle {\text{ }}x,y{\text{ odd}} }} {\frac{1}{{x!y!}}} = \frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} + \cdots + \frac{1}{{1!2009!}}$

I tried converting it into binomial coefficients and I'm getting sort of $\dfrac{2^{2009}}{2009!}$

Please help me.

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    @BrianM.Scott Thanks. I'm not sure how I messed that up.2012-06-30

2 Answers 2

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You’ve the right idea. First,

$\sum_{k=0}^{1004}\frac1{(2k+1)!(2010-2k-1)!}=\frac1{2010!}\sum_{k=0}^{1004}\binom{2010}{2k+1}\;.$

Now that last summation is simply the number of odd-sized subsets of a set of $2010$ elements. Since half the subsets of any non-empty set have odd cardinality, it’s simply $2^{2009}$. Thus, the desired sum is $\frac{2^{2009}}{2010!}\;.$

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By cancelling the even terms and doubling up the odd terms and dividing by $2$, the sum is $ \begin{align} &\frac{1}{2010!}\frac12\left(\sum_{k=0}^{2010}\binom{2010}{k}-\sum_{k=0}^{2010}(-1)^k\binom{2010}{k}\right)\\[6pt] &=\frac{1}{2010!}\frac12\left((1+1)^{2010}-(1-1)^{2010}\right)\\[6pt] &=\frac{2^{2009}}{2010!} \end{align} $