↑ stuart clark:
They are more ways for finding the closed from of your sum.
Skrytý text:It is well-known that the following integral representation holds true
Consequently,
![$
\sum_{k=1}^{2n-1}(-1)^{k-1}\cdot k\cdot\frac{1}{{2n\choose k}}
&=\sum_{k=1}^{2n-1}(-1)^{k-1}\cdot k\cdot (2n+1)\cdot\int_{0}^{1}x^k\cdot (1-x)^{2n-k}\mathrm dx\\[1em]
&=(2n+1)\cdot\int_{0}^{1}(1-x)^{2n}\cdot\sum_{k=1}^{2n-1}k\cdot (-1)^{k-1}\cdot\left (\frac{x}{1-x}\right )^k\mathrm dx\\[1em]
&=(2n+1)\cdot\int_{0}^{1}(1-x)^{2n}\cdot\left [(1-x)\cdot (2n-x)\cdot\left (\frac{x}{1-x}\right )^{2n}+x\cdot(1-x)\right ]\mathrm dx\\[1em]
&=(2n+1)\cdot\int_{0}^{1}\left [(1-x)^{2n}\cdot (2n-x)\cdot x^{2n}+x\cdot (1-x)^{2n+1}\right ]\mathrm dx\\[1em]
&=(2n+1)\cdot\left [2n\int_{0}^{1}x^{2n}\cdot (1-x)\mathrm dx-\int_{0}^{1}x^{2n+1}\cdot (1-x)\mathrm dx+\int_{0}^{1}x\cdot (1-x)^{2n+1}\mathrm dx\right ]\\[1em]
&=(2n+1)\cdot\left [2n\cdot\frac{1}{(2n+2)\cdot{2n+1\choose 2n}}-\frac{1}{{2n+2\choose 2n+1}}+\frac{1}{{2n+2\choose 1}}\right ].
$](/mathtex/c3/c3cf8130f3333f0b648fcd042f674ed9.gif "kopírovat do textarea")
After simplifications, one finally deduces that
as required. This concludes the proof.
Skrytý text:There is a really nice paper from A.M. Rockett (click
here to download it) devoted to the akin sum
Rockett develops an elemental method using recurrence relations. Generalizing his results, one can obtain similar recurrence relations for the parametric sum
Differentiating the last sum w.r.t. the variable
x, writing 2
n instead of
n and setting
x = -1, one arrives at the sum from the OP.
