# Find the sum of this series :$\frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} + \cdots + \frac{1}{{1!2009!}}$

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 $\dfrac{2^{2009}}{2009!}$

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Is the general term in the series $\dfrac{1}{{x!}{y!}}$ with $x+y=2010$ and $x$ odd ? – lhf Jun 30 '12 at 21:19
I too thought so. – Bazinga Jun 30 '12 at 21:19
@lhf Edited accordingly. – Pedro Tamaroff Jun 30 '12 at 21:23
Your answer $2^{2009}/2009!$ is correct. What problem are you having? – Logan Maingi Jun 30 '12 at 21:24
@Logan: It’s off by a factor of $2010$. – Brian M. Scott Jun 30 '12 at 21:29

$$\sum_{k=0}^{1004}\frac1{(2k+1)!(2010-2k-1)!}=\frac1{2010!}\sum_{k=0}^{1004}\binom{2010}{2k+1}\;.$$
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 $$\frac{2^{2009}}{2010!}\;.$$
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}