Is it true that if $\lim_{n\to\infty} n^\alpha a_n=+\infty$ for some $0<\alpha<1$ then $n(1-a_n/a_{n+1})\ge-\alpha$? Given a positive sequence $a_n $ convergent to $0$, is it true that if $\displaystyle \lim_{n\to\infty} n^\alpha a_n=+\infty$ for some $0<\alpha<1$ then $n(1-a_n/a_{n+1})\ge-\alpha$ for large enough $n$?
I've been trying to prove this for fun because it looked true but I haven't been able to
 A: The claim is not true. 
Pick 
$$a_n=\left\{
\begin{array}{lc}n^\beta &\mbox{ if } n \mbox{ is even } \\
n^{\gamma} &\mbox{ if } n \mbox{ is odd } \\
\end{array}
\right.$$
with $0 > \beta > \gamma >-\alpha$. Then
$$\lim_n n^\alpha a_n=\infty$$
But
$$2n(1-\frac{a_{2n}}{a_{2n+1}})\ge-\alpha \Leftrightarrow \\
2n(1-\frac{(2n)^\beta}{(2n+1)^\gamma})\ge-\alpha\Leftrightarrow \\
1-\frac{(2n)^\beta}{(2n+1)^\gamma}\ge-\frac{\alpha}{2n} \Leftrightarrow \\
1+\frac{\alpha}{2n} \ge\frac{(2n)^\beta}{(2n+1)^\gamma}
$$
Now, since $\beta> \gamma$ this cannot happen, as the LHS converges to 1, and the RHS goes to $\infty$.
A: The answer given by N.S. works just fine, I just wanted to point out that things are not even depended on $a$. There is a "global" counter-example that works for all $a$. 
EDIT: The following sequence lacks the "$\to0$ property. If anyone sees how it could be modified so it vanishes indeed but none of the other properties are lost, please feel free to comment.
Try $a_n=\frac{1}{\log(n)|\sin(n)|}$, for $n\geq 2$. For all $a\in(0,1)$, $\displaystyle{\frac{n^a}{\log(n)|\sin(n)|}\geq\frac{n^a}{\log(n)}\to\infty}$. But there is a subsequence of $\displaystyle{n(1-\frac{\log(n+1)|\sin(n+1)|}{\log(n)|\sin(n)|})}$ diverging to $-\infty$:
For the subsequence with indexes $n=2^m$ you have $\displaystyle{2^m(1-\frac{(m+1)\log(2)|\sin(2^{m+1})|}{m\log(2)|\sin(2^m)|})=2^m(1-(1+\frac{1}{m})2|\cos(2^m)|)}$. Now, as for the sequence $b_n=\cos(2^n)$: Note that from the identity $\cos(2x)=2\cos^2(x)-1$ we have that $b_n^2=2b_{n-1}^2-1$ for all $n$. By taking $\limsup$'s in the preceeding equality, we have the following: Let $s:=\limsup(b_n^2)$. Then $s=\max\{|\limsup(b_n)| , |\liminf(b_n)|\}$ and  $s^2=2s^2-1$ therefore $s=1$. Now s might be coming from the liminf or the limsup; In any case,  one can choose a subsequence of indexes -the indexes of the subsequence converging to limsup/ liminf-, namely $m_k$ such that eventually it will be $|\cos(2^{m_k})|\geq\frac{1}{2}$. For those $m_k$, for large $k$'s one has $$2^{m_k}(1-2(1+\frac{1}{m_k})|\cos(2^{m_k})|)\leq-\frac{2^{m_k}}{m_k}\to-\infty$$
