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How to prove the following inequality: $\forall n\geqslant 4:\dfrac {3^{n}+4^{n}+\cdots +\left( n+2\right) ^{n}} {\left( n+3\right) ^{n}} < 1$

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    It looks like the last term in the numerator should be $(n+2)^n$ and the denominator should be $(n+3)^n$. This is what Sasha answered.2012-07-18

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Induction base: verify for $n=4$: $ \underbrace{3^4 + 4^4 + 5^4 + 6^4}_{2258} < 7^4 = 2401 $

Induction step: Assuming inequality true for $n$, prove it true for $n+1$: $ \begin{eqnarray} \sum _{k=3}^{n+3} k^{n+1} &=& \sum _{k=3}^{n+2} k^{n+1}+(n+3)^{n+1} \leqslant (n+2)\sum _{k=3}^{n+2} k^{n}+(n+3)^{n+1} \\ &\stackrel{\text{by assumption}}{\leqslant}& (n+2) (n+3)^n+(n+3)^{n+1} \\ & < & 2 (n+3)^{n+1}<(n+4)^{n+1} \end{eqnarray} $ The last inequality follows because the sequence $a_n = \left( 1 + \frac{1}{n+3} \right)^{n+1}$ is increasing (proof of Brian M. Scott, given in comments): $\begin{eqnarray} \frac{a_{n+1}}{a_n} &=& \left(1+\frac{1}{n+4}\right) \left( 1 - \frac{1}{(n+4)^2} \right)^{n+1} \\ &\stackrel{\text{binomial theorem}}{>}& \left(1+\frac{1}{n+4}\right) \left( 1 - \frac{n+1}{(n+4)^2} \right) = 1 + \frac{2n+11}{(n+4)^3} > 1 \end{eqnarray}$

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    @BrianM.Sco$t$$t$ Thanks for fixing my proof. I have edited to include your proof, and turned the answer into CW.2012-07-19