Question

Show that the sequence En=(1+1/n)^n is bounded and increasing

Answer 1

$e_n=\left( 1+\frac{1}{n} \right) ^n\\\left( 1+\frac{1}{n} \right) ^n=\sum_{i=0}^n{C_{n}^{i}\left( \frac{1}{n} \right) ^{n-i}}=1+n\cdot \frac{1}{n}+\frac{n\left( n-1 \right)}{2n^2}+\frac{n\left( n-1 \right) \left( n-2 \right)}{2\cdot 3n^3}+...+\frac{n!}{n!n^n}=\\=2+\frac{1}{2!}\frac{n-1}{n}+\frac{1}{3!}\frac{n-2}{n}\frac{n-1}{n}+...+\frac{1}{n!}\frac{1}{n}\frac{2}{n}...\frac{n}{n}<\\<2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}<2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}<2+\frac{1/2}{1-1/2}=3\\Bounded\,\,by\,\,3\\Next,\\e_n=2+\frac{1}{2!}\left( 1-\frac{1}{n} \right) +\frac{1}{3!}\left( 1-\frac{1}{n} \right) \left( 1-\frac{2}{n} \right) +...+\frac{1}{n!}\left( 1-\frac{1}{n} \right) \left( 1-\frac{2}{n} \right) ...\left( 1-\frac{n-1}{n} \right) \\e_{n+1}=2+\frac{1}{2!}\left( 1-\frac{1}{n+1} \right) +\frac{1}{3!}\left( 1-\frac{1}{n+1} \right) \left( 1-\frac{2}{n+1} \right) +...+\frac{1}{n!}\left( 1-\frac{1}{n+1} \right) \left( 1-\frac{2}{n+1} \right) ...\left( 1-\frac{n-1}{n+1} \right) +\\+\frac{1}{\left( n+1 \right) !}\left( 1-\frac{1}{n+1} \right) \left( 1-\frac{2}{n+1} \right) ...\left( 1-\frac{n}{n+1} \right) >\\>2+\frac{1}{2!}\left( 1-\frac{1}{n} \right) +\frac{1}{3!}\left( 1-\frac{1}{n} \right) \left( 1-\frac{2}{n} \right) +...+\frac{1}{n!}\left( 1-\frac{1}{n} \right) \left( 1-\frac{2}{n} \right) ...\left( 1-\frac{n-1}{n} \right) =e_n\\Incre\mathrm{a}\sin g\\$