# inequality with sum of powers

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. – Ross Millikan Jul 18 '12 at 21:52

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): $$\begin{eqnarray} \frac{a_{n+1}}{a_n} &=& \left(1+\frac{1}{n+4}\right) \left( 1 - \frac{1}{(n+4)^2} \right)^{n+1} \\ &\stackrel{\text{binomial theorem}}{>}& \left(1+\frac{1}{n+4}\right) \left( 1 - \frac{n+1}{(n+4)^2} \right) = 1 + \frac{2n+11}{(n+4)^3} > 1 \end{eqnarray}$$
The proof of your last statement is wrong: $\dfrac {n+5} {n+4}\left( 1+\dfrac {2} {n+3}\right) ^{n+1}=a_{n+1}a_{n}$ $\neq \dfrac {a_{n+1}} {a_{n}}$ ! – Maria Mikolayevskaya Jul 19 '12 at 6:42
Maria M.’s objection is correct: $$\frac{a_{n+1}}{a_n}=\left(1+\frac1{n+4}\right)^{n+2}\left(1-\frac1{n+4}\right)‌​^{n+1}\;.$$ – Brian M. Scott Jul 19 '12 at 8:20
However, $$\frac{a_{n+1}}{a_n}=\left(1+\frac1{n+4}\right)\left(1-\frac1{(n+4)^2}\right)^{‌​n+1}\ge \left(1+\frac1{n+4}\right)\left(1-\frac{n+1}{(n+4)^2}\right)$$ by the binomial theorem, and $$\left(1+\frac1{n+4}\right)\left(1-\frac{n+1}{(n+4)^2}\right)=\frac{3(n+5)}{(n+‌​4)^2}>0\;,$$ so the sequence is indeed increasing. – Brian M. Scott Jul 19 '12 at 8:43