# 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,

 $\left(1+{1\over n}\right)^{n}<{n\over n+1}\left(1+{1\over m}\right)^{m+1}$
Title proof of bounds for e ProofOfBoundsForE 2013-03-22 15:48:51 2013-03-22 15:48:51 rspuzio (6075) rspuzio (6075) 4 rspuzio (6075) Proof msc 33B99