How can show
$$\frac{{{{\left( {1 + \frac{1}{n}} \right)}^{{n^2}}}}}{{{{\left( {1 + \frac{1}{{n + 1}}} \right)}^{{{(n + 1)}^2}}}}} < \frac{1}{{{{\left( {1 + \frac{1}{n}} \right)}^n}}}$$
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Ron Gordon
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user83783
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1This perhaps works, take the ln on both sides and write fractions inside with a common denominator. Keep splitting the ln terms. Work out (n+1)² and distribute over ln terms. It clears a lot up – imranfat Oct 15 '13 at 17:42
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Multiply both sides by the inverse of the RHS and note the difference in the remaining exponents on the LHS. – abiessu Oct 15 '13 at 17:45
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Do you already know that $\left(1+\frac{1}{k}\right)^k$ is increasing? Your inequality is an immediate consequence. – André Nicolas Oct 15 '13 at 18:01
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Using Bernoulli's Inequality, $$ \begin{align} \frac{\left(1+\frac1{n+1}\right)^{(n+1)^2}}{\left(1+\frac1n\right)^{n^2}} &=\left(\frac{n(n+2)}{(n+1)^2}\right)^{(n+1)^2}\left(1+\frac1n\right)^{2n+1}\\ &=\left(1-\frac1{(n+1)^2}\right)^{(n+1)^2}\left(1+\frac1n\right)^{n+1}\left(1+\frac1n\right)^n\\ &\ge\left(1-\frac1{n+1}\right)^{n +1}\left(1+\frac1n\right)^{n+1}\left(1+\frac1n\right)^n\\ &=\left(1+\frac1n\right)^n \end{align} $$
Second approach: Using this result $$ \left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1} $$ Raise both sides to the $n+1^\text{st}$ power $$ \left(1+\frac1n\right)^{n(n+1)}\le\left(1+\frac1{n+1}\right)^{(n+1)^2} $$ Divide both sides by $\left(1+\frac1n\right)^n\left(1+\frac1{n+1}\right)^{(n+1)^2}$ $$ \frac{\left(1+\frac1n\right)^{n^2}}{\left(1+\frac1{n+1}\right)^{(n+1)^2}}\le\frac1{\left(1+\frac1n\right)^n} $$
