MHB Prove an identity with binomial coefficients

AI Thread Summary
The discussion centers on proving the identity involving binomial coefficients: $\sum_{j=1}^{2n-1}\frac{(-1)^{j-1}j}{{2n \choose j }} = \frac{n}{n+1}$. Participants express difficulty in using proof by induction and seek alternative approaches. Observations highlight that for small values of n, the left side simplifies intriguingly, yet the overall problem remains complex. A referenced paper on alternating sums of binomial coefficients may provide additional insights, though its notation is dense. The conversation reflects a shared acknowledgment of the challenge posed by the identity.
lfdahl
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Prove, that

$\sum_{j=1}^{2n-1}\frac{(-1)^{j-1}j}{{2n \choose j }} = \frac{n}{n+1}$

i have tried with proof by induction, but it is very difficult to use this technique.

I should be very glad to see any approach, that can crack this nut.
 
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lfdahl said:
Prove, that

$\sum_{j=1}^{2n-1}\frac{(-1)^{j-1}j}{{2n \choose j }} = \frac{n}{n+1}$

i have tried with proof by induction, but it is very difficult to use this technique.

I should be very glad to see any approach, that can crack this nut.
This looks like a difficult problem. The only help I could find online is a paper Alternating sums of the reciprocals of binomial coefficients. You might find that helpful if you can wade your way through the dense notation.

It is fascinating to see how the formula $\sum_{j=1}^{2n-1}\frac{(-1)^{j-1}j}{{2n \choose j }} = \frac{n}{n+1}$ works for small values of $n$. For example, when $n=4$ the left side is $\frac18 - \frac2{28} + \frac3{56} - \frac4{70} + \frac5{56} - \frac6{28} + \frac78$, which magically simplifies to $\frac45.$ Notice that if you pair together the terms from each end of the sum that have the same denominator, their sum has the constant numerator $2n$ (except that the middle term stands on its own, and its numerator is $n$). But that does not seem to make the problem any simpler.
 
sorry about my post, absolute nonsense mate
 
Fermat said:
sorry about my post, absolute nonsense mate
No problem at all! ;)

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Opalg said:
This looks like a difficult problem. The only help I could find online is a paper Alternating sums of the reciprocals of binomial coefficients. You might find that helpful if you can wade your way through the dense notation.

It is fascinating to see how the formula $\sum_{j=1}^{2n-1}\frac{(-1)^{j-1}j}{{2n \choose j }} = \frac{n}{n+1}$ works for small values of $n$. For example, when $n=4$ the left side is $\frac18 - \frac2{28} + \frac3{56} - \frac4{70} + \frac5{56} - \frac6{28} + \frac78$, which magically simplifies to $\frac45.$ Notice that if you pair together the terms from each end of the sum that have the same denominator, their sum has the constant numerator $2n$ (except that the middle term stands on its own, and its numerator is $n$). But that does not seem to make the problem any simpler.

Thanks a lot, Opalg, for your thorough considerations and numerical observations. I did observe my self, that pairing the terms from each end of the sum results in a common denominator ($2n$), but I was not able to proceed and get any closer to a final result ...:(
Well, anyway, from your considerations its a "comfort" to know, that the problem in fact is difficult, and that I didn´t overlook any obvious approach. Thankyou once again!
 
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