strange inequality involving infinite series while doing a complex analysis exercise, i came to a strange inequality which i don't know how to interpretate. Suppose you have a sequence $\{a_j\}$ of positive real number. Let $\rho$ a positive real number. The inequality i found after some calculation is
$$\sum_{j=1}^{+\infty}\frac{1}{|a_j|^{\rho +\epsilon}}\leq \sum_{j=1}^{+\infty}\frac{1}{|a_j|^{\rho-\epsilon}}$$
for every $\epsilon>0$.
My question is: can i deduce something from this inequality? for example the convergence of the first series (that with $+\epsilon$)? Can i deduce nothing? Is that inequality surely false?Is it always true, so that i can't deduce nothing in particular?
EDIT: the sequence $a_j$ tends to $\infty$
 A: If $|a_j|\geqslant1$ for every $j$ then $|a_j|^{\rho+\epsilon}\geqslant|a_j|^{\rho-\epsilon}$ hence indeed,
$$
\sum_{j=1}^{+\infty}\frac1{|a_j|^{\rho+\epsilon}}\leqslant\sum_{j=1}^{+\infty}\frac1{|a_j|^{\rho-\epsilon}}.
$$
Otherwise no comparison holds (consider the limit $|a_1|\to0$, every other $a_j$ fixed).
But of course, if the series $\displaystyle\sum_j\frac1{|a_j|^{\rho-\epsilon}}$ converges, then $|a_j|\geqslant1$ for every $j$ large enough hence the series $\displaystyle\sum_j\frac1{|a_j|^{\rho+\epsilon}}$ converges as well.
A: Basicaly, what you are saying is that if $a_j>0$ and $\alpha<\beta$, then
$$
\sum_{j=1}^\infty\frac{1}{a_j^\beta}\le\sum_{j=1}^\infty\frac{1}{a_j^\alpha}.
$$
This is certainly true if $a_j\ge1$. Since in your case $a_j\to\infty$, this holds for all $j$ large enough. But the inequality is not true in general. Let $N\in\mathbb{N}$ and define
$$
a_j=\begin{cases}
1/2 & \text{if }j\le N,\\
2^{N-j} & \text{if }j< N.
\end{cases}
$$
Then
$$
\sum_{j=1}^\infty\frac{1}{a_j^\beta}-\sum_{j=1}^\infty\frac{1}{a_j^\alpha}=
N(2^\beta-2^\alpha)+\frac{1}{1-2^{-\beta}}-\frac{1}{1-2^{-\alpha}},
$$
which is positive if $N$ is large enough.
In general there is nothing you can deduce from the inequality, since the right hand side can be $\infty$ and the left hand side $<\infty$.
