Multiplying and dividing, we have
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(1+1n)n=(1+1m)mn∏k=m+1(1+1k)k(1+1k-1)k-1 |
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As was shown in the parent entry, the quotients in the product can
be simplified to give
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(1+1n)n=(1+1m)mn∏k=m+1(1-1k2)k(1+1k-1) |
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By the inequality for differences of powers,
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(1-1k2)k<1-kk2+k-1=(k+1)(k-1)k2+k-1 |
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Hence, we have the following upper bound:
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(1+1n)n<(1+1m)mn∏k=m+1k2+kk2+k-1 |
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By cross mutliplying, it is easy to see that
and, hence,
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(1+1n)n<(1+1m)mn∏k=m+1k2k2-1. |
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Factoring the rational function in the product, terms cancel and we
have
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n∏k=m+1k2(k+1)(k-1)=n(m+1)(n+1)m=nn+1(1+1m) |
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Combining,