Is $2 \sum_{n=m+2}^{\infty}\dfrac{1}{n^{\alpha}} \geq \sum_{n=m+1}^{\infty}\dfrac{1}{n^{\alpha}}$? Working in  a project I came across to the following problem, to investigate if the following  inequality is true for $\alpha>2,$ and $m$ a positive integer:
                   $$2 \sum_{n=m+2}^{\infty}\dfrac{1}{n^{\alpha}} \geq \sum_{n=m+1}^{\infty}\dfrac{1}{n^{\alpha}}.$$
I have made some numerical experiments in Worfram which indicates that this is true, but I have no idea how to perform ao proof.
 A: This is equivalent to just asking the question of whether
$$\frac{1}{(m + 1)^{\alpha}} \le \sum_{n = m + 2}^{\infty} \frac{1}{n^{\alpha}}$$
or equivalently, whether
$$\sum_{k = 2}^{\infty} \left(\frac{m + 1}{m + k}\right)^{\alpha} \ge 1$$
Now fix an exponent $\alpha$. The summands are each increasing in $m$, tending towards $1$ as $m \to \infty$ (which immediately implies the desired inequality for $m$ large enough, depending on $\alpha$); therefore, it is sufficient to study $m = 1$. We need to show that
$$2^{\alpha} \sum_{j = 3}^{\infty} \frac{1}{j^{\alpha}} \ge 1.$$
However, if $\alpha$ is sufficiently large this does not hold: in fact, 
$$\lim_{\alpha \to \infty} 2^{\alpha} \sum_{j = 3}^{\infty} \frac{1}{j^{\alpha}} = 0.$$

So the punchline is that for each fixed $m$, this works if $\alpha$ is small enough. And for each fixed $\alpha$, it works if $m$ is large enough.
A: When $\alpha = 4$ and $m=1$ you can do the sums in closed form:
$$
2 \sum_{n=3}^\infty \frac{1}{n^4} - \sum_{n=2}^\infty \frac{1}{n^4} 
= \frac{\pi^4}{90}-\frac98 < 0
$$
