Inequality for an alternating sum of binomial coefficients times a fractionI need to show that ∑ n = 0 x ( − 1 ) n ( x n ) ( 1 n + l − 1 − 1 n + l ) &gt; 0 ,for any l &gt; 0. I tried to prove this, but I didn't get anywhere.Apparently the above sum is equal to ( l − 2 ) ! ( x + 1 ) ! ( x + l ) ! . However I have no idea how to prove this either. Any tips on how to approach this would be greatly appreciated!

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Inequality for an alternating sum of binomial coefficients times a fractionI need to show that      ∑          n      =      0        x    (  −  1      )    n                                (                            x        n                              )                          (          1              n        +        l        −        1              −          1              n        +        l              )    &amp;gt;  0  ,for any   l  &amp;gt;  0. I tried to prove this, but I didn&#039;t get anywhere.Apparently the above sum is equal to             (      l      −      2      )      !      (      x      +      1      )      !              (      x      +      l      )      !      . However I have no idea how to prove this either. Any tips on how to approach this would be greatly appreciated!

Megan Mathis · Accepted Answer

Since       1    a    =      ∫          0              1            z          a      −      1          d  z, we have:                              ∑                      n            =            0                                x                                                              (                                            x            n                                              )                                      (        −        1                  )          n                          (                      1                          n              +              l              −              1                                −                      1                          n              +              l                                )                                    =                              ∫                      0                                1                                    ∑                      n            =            0                                x                                                              (                                            x            n                                              )                                      (        −        1                  )          n                          (                      z                          n              +              l              −              2                                −                      z                          n              +              l              −              1                                )                        d        z                                          =                              ∫                      0                                1                          (        1        −        z                  )          x                          (                      z                          l              −              2                                −                      z                          l              −              1                                )                        d        z                                          =                              ∫                      0                                1                          (        1        −        z                  )                      x            +            1                                    z                      l            −            2                                  d        z        .            The last integral is clearly positive and can be computed through Euler&#039;s beta function, or simply by integration by parts.

Inequality for an alternating sum of binomial coefficients times a fraction I need to show that

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