We have to prove
i=1∑ni1≤n!1i=2∏n(i−1)+2i=2∑ni1
Consider
i=1∑ni1=1+i=2∑ni1 Subtract (2i=2∑ni1) from both sides
i=1∑ni1−2i=2∑ni1=1+i=2∑ni1−2i=2∑ni1
i=1∑ni1−2i=2∑ni1=1−i=2∑ni1
Hence we can rewrite the original inequality as
1−i=2∑ni1≤n!1i=2∏n(i−1) By the Weierstrass inequality (since i1∈(0,1] for all i∈Z+) we have
1−i=2∑ni1≤i=2∏n(1−i1)=i=2∏n(ii−1)=n!1i=2∏n(i−1) For all i∈Z+
i−i−1=i+i−1i−(i−1)≤1
Then
i−1≤i−1 Hence
1−i=2∑ni1≤n!1i=2∏n(i−1)≤n!1i=2∏n(i−1) Therefore, we prove that
i=1∑ni1≤n!1i=2∏n(i−1)+2i=2∑ni1
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