proof of bounds for e

Multiplying and dividing, we have

\\left(1+{1\\over n}\\right)^{n}=\\left(1+{1\\over m}\\right)^{m}\\prod_{k=m+1}^{n}{%\\left(1+{1\\over k}\\right)^{k}\\over\\left(1+{1\\over k-1}\\right)^{k-1}}

As was shown in the parent entry, the quotients in the product canbe simplified to give

\\left(1+{1\\over n}\\right)^{n}=\\left(1+{1\\over m}\\right)^{m}\\prod_{k=m+1}^{n}%\\left(1-{1\\over k^{2}}\\right)^{k}\\left(1+{1\\over k-1}\\right)

By the inequality for differences of powers,

\\left(1-{1\\over k^{2}}\\right)^{k}<1-{k\\over k^{2}+k-1}={(k+1)(k-1)\\over k^{2}+%k-1}

Hence, we have the following upper bound:

\\left(1+{1\\over n}\\right)^{n}<\\left(1+{1\\over m}\\right)^{m}\\prod_{k=m+1}^{n}{k%^{2}+k\\over k^{2}+k-1}

By cross mutliplying, it is easy to see that

{k^{2}+k\\over k^{2}+k-1}\\leq{k^{2}\\over k^{2}-1}

and, hence,

\\left(1+{1\\over n}\\right)^{n}<\\left(1+{1\\over m}\\right)^{m}\\prod_{k=m+1}^{n}{k%^{2}\\over k^{2}-1}.

Factoring the rational function in the product, terms cancel and wehave

\\prod_{k=m+1}^{n}{k^{2}\\over(k+1)(k-1)}={n(m+1)\\over(n+1)m}={n\\over n+1}\\left(%1+{1\\over m}\\right)

Combining,

\\left(1+{1\\over n}\\right)^{n}<{n\\over n+1}\\left(1+{1\\over m}\\right)^{m+1}