When $n=2$, it directly works.
Assume it works for $n=k\ge 2$, and try to show that for $n=k+1$.
$1+...+\frac{1}{k^2}>\frac{3k}{2k+1}$ is given, so it suffices to show $\frac{1}{(k+1)^2}\ge\frac{3(k+1)}{2(k+1)+1}-\frac{3k}{2k+1}=\frac{3}{4(k+1)^2-1}\Leftrightarrow (k+1)^2\ge 1$, which is obvious.
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 can prove by induction that $$(1-x_1)(1-x_2)\cdots (1-x_n)\geq 1-x_1-x_2-\ldots -x_n$$ for all real numbers $x_1,x_2,\ldots,x_n\in[0,1]$ (the equality holds, by the way, iff at least $n-1$ of $x_1,x_2,\ldots,x_n$ are $0$). Using that inequality, we have $$\frac{1}{\left(1+\frac1{2n}\right)^n}=\left(1-\frac{1}{2n+1}\right)^n\geq 1-\frac{n}{2n+1}>\frac{1}{2}.$$