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 can be 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)

\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 we have

\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}