Prove that (1+1n)n<10!+11!+12!+...+1n!
for n>1,n∈N.
Answer
We have by the binomial identity that
(1+1n)n=n∑k=0(nk)1nk=n∑k=0n!(n−k)!nk⋅1k!=n∑k=0n⋅(n−1)⋯(n−k+1)n⋅n⋯n⋅1k!now the first factor is <1 for k≥2<n∑k=01k!
for n≥2.
Prove that (1+1n)n<10!+11!+12!+...+1n!
Answer
We have by the binomial identity that
(1+1n)n=n∑k=0(nk)1nk=n∑k=0n!(n−k)!nk⋅1k!=n∑k=0n⋅(n−1)⋯(n−k+1)n⋅n⋯n⋅1k!now the first factor is <1 for k≥2<n∑k=01k!
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