Growing slower than exponential imply some divergence

Question

Is the following true?

Let $(x_n)_{\mathbb{N}}$ be a sequence such that for all $c > 1$, exists $m_c \in \mathbb{N}$ whith $x_n < c^n, \forall n > m_c$. Then exists $\gamma > 0$, for which

$$\sum_{n\geq 1} \frac{1}{x_n^{\gamma}} = \infty.$$

Answer 1

No, this is false. For each $N\in\mathbb{N}$, choose $m(N)\in \mathbb{N}$ such that $$\sum_{n\geq m(N)}\frac{1}{(1+1/N)^{n/N}}<1/2^N$$ (such an $m(N)$ exists since the sum converges). Now let $x_n=(1+1/N)^n$ for the greatest $N$ such that $n\geq m(N)$ (or $x_n=1$ if $n\geq m(N)$ for all $N$, but it is easy to see that can't actually happen). Since the $N$ used to define $n$ grows as $n$ grows, it is clear that $x_n$ is eventually less than $c^n$ for any fixed $c>1$. However, for any $\gamma<0$, let $N$ be such that $1/N<\gamma$. Then $$\sum_{n\geq m(N)}\frac{1}{x_n^\gamma}\leq\sum_{M\geq N}\sum_{n\geq m(M)}\frac{1}{(1+1/M)^{n\gamma}}\leq \sum_{M\geq N}\sum_{n\geq m(M)}\frac{1}{(1+1/M)^{n/M}}<\sum_{M\geq N}\frac{1}{2^M}<\infty$$ (the first inequality is because every $x_n^\gamma$ for $n\geq m(N)$ is equal to $(1+1/M)^{n\gamma}$ for some $M\geq N$, namely the greatest $M$ such that $n\geq m(M)$). It follows that $$\sum_{n\geq 1}\frac{1}{x_n^\gamma}<\infty$$ as well.

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This sort of "diagonalization" argument comes up all the time in questions like this and is very flexible. In general, if you have countably many sequences, and you want to construct a single sequence that is eventually smaller (or larger) that each one of them but shares certain convergence properties with all of them, you can typically do so by taking a sequence that agrees with each one for "long enough" to get the required convergence before moving on to the next one.