TITLE: Find the sum of this series :$ \frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} +  \cdots  + \frac{1}{{1!2009!}}$
QUESTION [8 upvotes]: 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.

REPLY [8 votes]: By cancelling the even terms and doubling up the odd terms and dividing by $2$, the sum is
$$
\begin{align}
&\frac{1}{2010!}\frac12\left(\sum_{k=0}^{2010}\binom{2010}{k}-\sum_{k=0}^{2010}(-1)^k\binom{2010}{k}\right)\\[6pt]
&=\frac{1}{2010!}\frac12\left((1+1)^{2010}-(1-1)^{2010}\right)\\[6pt]
&=\frac{2^{2009}}{2010!}
\end{align}
$$