Fix $m$. From the base case $n=0$, assume that $(m+1)^n>mn$ holds for some $n\in\Bbb N$. Then for the inductive step of $n+1$, we have $$(m+1)^{n+1}=(m+1)(m+1)^n>(m+1)mn>m(n+1)\impliedby m+1>1+\frac1n$$ which is true for all $m>0$. When $m=0$, we have $(0+1)^n>0\cdot n$ which is also true, so induction on $n$ is complete.
Fix $n$. From the bases case $m=0$, assume that $(m+1)^n>mn$ holds for some $m\in\Bbb N$. Then for the inductive step of $m+1$, we have \begin{align}(m+1+1)^n&=(m+1)^n+\binom n1(m+1)^{n-1}+\cdots+\binom n{n-1}(m+1)+1\\&>mn+n(m+1)^{n-1}+\cdots+n(m+1)+1\\&=(m+1)n+n(m+1)^{n-1}+\cdots+n+1>(m+1)n\end{align} which is true for all $n\ge0$. Induction on $m$ is complete, and the result follows.
Thanks to the help of @maxmilgram, I've managed to figure out the solution.
SOLUTION:
- Test with n = 1, which gives:
$$0\le1\le\frac{127}{7}, \text{which is true}$$
- Assumption:
$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$
- Induction step:
$$\frac{127}{7}\cdot\ n^7\le\sum_{k=n+1}^{2n+1}k^6\le\frac{127}{7}\cdot\ (n+1)^7$$
And since:
$$\sum_{k=n+1}^{2n+1}k^6=\sum_{k=n}^{2n-1}k^6-n^6+(2n)^6+(2n+1)^6$$
we can write the induction step in terms of the assumption:
$$\frac{127}{7}n^7\le\sum_{k=n}^{2n-1}k^6-n^6+\left(2n\right)^6+\left(2n+1\right)^6\le\frac{127}{7}\left(n+7\right)^7$$
$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \sum _{k=n}^{2n-1}k^6\le \frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6$$
In other words, if we could prove that:
$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \frac{127}{7}\left(n-1\right)^7$$ and
$$\frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\ge \frac{127}{7}n^7$$
we would know that the induction step is indeed true. After some simplification we end up with:
$$-573n^5+395n^4-795n^3+321n^2-139n+\frac{120}{7}\le 0, \text{which is true since } n\ge 1$$
$$189n^5+395n^4+475n^3+321n^2+115n+\frac{120}{7}\ge 0, \text{which is true since } n \ge 1$$
In other words, we have shown that the induction step is true if the induction assumption also is true. Therefore, the statement is true for all positive integers n.
Best Answer
You also have that $\sin x\ge0$ and from $\cos kx\le1$ you conclude that $$ \sin x\cos kx\le\sin x $$ So this is the easy step.
The issue is to prove $\sin kx\cos x\le k\sin x$, where your argument is faulty. But this is not a real problem: if $\sin kx\le0$, then $\sin kx\cos x\le0\le k\sin x$. If $\sin kx>0$, then from $\cos x\le 1$ you obtain $\sin kx\cos x\le \sin kx$ and you can apply the induction hypothesis.