How to prove the following equality? ∑n=1∞1∏k=1m(n+k)=1(m−1)m!

Question

How to prove the following equality?  ∑n=1∞1∏k=1m(n+k)=1(m−1)m!

Deegan Mullen · Accepted Answer

∑n=1∞1∏k=1m(n+k)=∑n=1∞n!(n+m)!=1m!∑n=1∞1(m+nn)  Now  1(m+nn)=m∫01dxxm−1(1−x)n  so that the sum in question is equal to, upon reversing sum and integral  m∫01dxxm−1∑n=1∞(1−x)n=m∫01dxxm−11−xx  =m∫01dx(xm−2−xm−1)  =m(1m−1−1m)  =1m−1  Putting this all together,  ∑n=1∞1∏k=1m(n+k)=1(m−1)m!  as was to be shown.

kumewekwah0 · Accepted Answer

Note that ∏k=1m(n+k)=(n+m)!n!, then prove by induction the explicit formula for the s-th partial sum  ∑n=1s1∏k=1m(n+k)=∑n=1sn!(n+m)!  =1m−1(1m!−(s+1)!(s+m)!  and take the limit s→∞

RizerMix · Accepted Answer

The partial fraction identity at the top can be shown by induction on m. The case m=1 is trivial. The induction step gives 1(m−1)!∑k=0m−1(−1)k((m−1),(k))1n+k+11n+m+1 =1(m−1)! ∑k=0m−1(−1)k((m−1),(k))1m−k(1n+k+1−1n+m+1 The first part of the sum is 1(m−1)!∑k=0m−1(−1)k((m−1),(k))1m−k1n+k+1=1m! ∑k=0m−1(−1)k((m),(k))1n+k+1 The second part is −1n+m+11(m−1)!∑k=0m−1(−1)k((m−1),(k))1m−k=−1n+m+11m!∑k=0m−1(−1)k((m),(k)) =−1n+m+11m!(0−(−1)m) =1n+m+1(−1)m1m! Putting these two together completes the induction.

How to prove the following equality? \sum_{n=1}^\infty\frac{1}{\prod_{k=1}^m(n+k)}=\frac{1}{(m-1)m!}

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