Prove an identity with binomial coefficients

[lfdahl]

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.

 

[Fermat1]

Isn't it basically this. Must be if RHS is the same

Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ - Mathematics Stack Exchange

 

[Opalg]

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.

 

[Fermat1]

sorry about my post, absolute nonsense mate

 

[lfdahl]

sorry about my post, absolute nonsense mate

No problem at all! ;)

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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!