$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1)$ is true for any $n\in\mathbb N$? Question : Is the following true for any $n\in\mathbb N$?
$$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1).$$
Note that $\lfloor x\rfloor$ is the largest integer not greater than $x$. 
Motivation : I've been able to prove the following $(\star)$: 
"For any $n\in\mathbb N$,
$$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^2}\right)^{-1}\right\rfloor=n-1."$$
Then, I reached the above expectation. However, I can't prove that this expectation is true. Can anyone help?
Proof for $(\star)$ : The $n=1$ case is obvious. In the following, let $n\ge 2$. First, since 
$$\sum_{k=n}^{\infty}\frac {1}{k^2}\gt\sum_{k=n}^{\infty}\frac{1}{k(k+1)}=\sum_{k=n}^{\infty}\left(\frac 1k-\frac{1}{k+1}\right)=\frac 1n,$$
we get
$$\begin{align}\sum_{k=n}^{\infty}\frac{1}{k^2}\gt \frac 1n\qquad(1)\end{align}$$
Next, since 
$$\sum_{k=n}^{\infty}\frac {1}{k^2}\lt\sum_{k=n}^{\infty}\frac{1}{k(k-1)}=\sum_{k=n}^{\infty}\left(\frac 1{k-1}-\frac{1}{k}\right)=\frac 1{n-1},$$
we get
$$\begin{align}\sum_{k=n}^{\infty}\frac{1}{k^2}\lt \frac 1{n-1}\qquad(2)\end{align}$$
$(1)(2)$ give us
$$\frac 1n\lt\sum_{k=n}^{\infty}\frac{1}{k^2}\lt\frac{1}{n-1}\rightarrow n-1\lt\left(\sum_{k=n}^{\infty}\frac{1}{k^2}\right)^{-1}\lt n.$$
Hence, $$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^2}\right)^{-1}\right\rfloor=n-1$$
as desired.

Yes, it is true:
$$\sum_{k=n}^{\infty} \frac{1}{k^3} < \sum_{k=n}^{\infty} \frac{1}{k(k^2-1)} = \sum_{k=n}^{\infty} \Big( \frac{1}{2k(k-1)} - \frac{1}{2k(k+1)}\Big) = \frac{1}{2n(n-1)}$$ and $$\sum_{k=n}^{\infty} \frac{1}{k^3} > \sum_{k=n}^{\infty} \frac{k}{k^4 + \frac{1}{4}} = \sum_{k=n}^{\infty} \Big( \frac{1}{2k^2 -2k + 1} - \frac{1}{2(k+1)^2 - 2(k+1) + 1}\Big)$$ $$=\frac{1}{2n^2 - 2n + 1} = \frac{1}{2n(n-1) + 1}.$$
