Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
Your answer $2^{2009}/2009!$ is correct. What problem are you having? –  Logan Maingi Jun 30 '12 at 21:24
2  
@Logan: It’s off by a factor of $2010$. –  Brian M. Scott Jun 30 '12 at 21:29
show 1 more comment

2 Answers

up vote 10 down vote accepted

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 $$\frac{2^{2009}}{2010!}\;.$$

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.