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Question-208322




Question Number 208322 by alcohol last updated on 11/Jun/24
Answered by Berbere last updated on 11/Jun/24
Π_(k=1) ^(n−1) (1+(1/k))^k =Π_(k=1) ^(n−1) (((1+k)^k )/k^k )=Π_(k=1) ^(n−1) (((1+k)^(k+1) )/((k+1).k^k ))  =Π_(k=1) ^(n−1) (1/((k+1))).Π_(k=1) ^(n−1) (((1+k)^(k+1) )/k^k )=(1/(n!)).n^n =(n^(n−1) /((n−1)!))
$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)^{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\left(\mathrm{1}+{k}\right)^{{k}} }{{k}^{{k}} }=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\left(\mathrm{1}+{k}\right)^{{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right).{k}^{{k}} } \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)}.\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\left(\mathrm{1}+{k}\right)^{{k}+\mathrm{1}} }{{k}^{{k}} }=\frac{\mathrm{1}}{{n}!}.{n}^{{n}} =\frac{{n}^{{n}−\mathrm{1}} }{\left({n}−\mathrm{1}\right)!} \\ $$
Answered by MathematicalUser2357 last updated on 13/Jun/24
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