Convergence series

Question

Suppose for a fixed $s$, $(a_k)$ is a given sequence with $a_k>0$, $\sum_{k=1}^\infty a_kk^s<\infty$, but for any $\epsilon>0$, $\sum_{k=1}^\infty a_kk^{s+\epsilon}=\infty$. Can we conclude that there exists a $b>0$ such that $\sum_{k=1}^\infty a_kk^{s}(\log k)^b=\infty$?

Answer 1

No. I think there are counterexamples to all claims of this sort.

Let's consider only sequences of nonnegative reals for simplicity. Suppose that you have two countable families of sequences: For each $i \in \mathbb{N}$ you have a sequence $(b_{i,k})_{k \in \mathbb{N}}$ and a sequence $(c_{i,k})_{k \in \mathbb{N}}$. Suppose that $b_i = o(c_j)$ for all $i, j$ where $x = o(y)$ for two sequences means $\lim_k x_k/y_k = 0$ for all large enough $k$. Suppose that at least one of the sequences $b_i$ is not eventually $0$, so that for every $i$, $c_{i,j} > 0$ for infinitely many $j$.

Then the following holds: There exists $(a_k)_{k \in \mathbb{N}}$ such that $\sum_k a_k b_{j,k} < \infty$ and $\sum_k a_k c_{j,k} = \infty$ for all $j$.

To see this, pick a sequence $n_0 < m_0 < n_1 < m_1 < n_2 < m_2 ...$, such that $b_{i,n} \leq 2^{-k} c_{j,n}$ for every $i, j \leq k$ and $n \geq n_k$, and $c_{t(k),m_k} > 0$, where $t : \mathbb{N} \to \mathbb{N}$ is some fixed surjective $\infty$-to-$1$ function with $t(k) \leq k$ for all $k$ (e.g. the function defined by $t(0) = 0$, $t(2^k m) = k$ for odd $m$).

Now for all $k$ define $a_{m_k} = 1/c_{t(k),m_k}$, and $a_n = 0$ for all $n \notin \{m_k \;|\; k \in \mathbb{N}\}$.

Consider any $i \in \mathbb{N}$. From some point on, $m_k > i$ for all $k$, so for large $k$, $a_{m_k} b_{i,m_k} = b_{i,m_k}/c_{t(k),m_k} \leq 2^{-k}$, and thus $\sum_k a_{m_k} b_{i,m_k} < \infty$ since it is eventually majored by a geometric series.

On the other hand, fixing again some $i \in \mathbb{N}$, by the choice of $t$, there are infinitely many $s_0, s_1, ...$ such that $t(s_j) = i$, so \begin{align} \sum_k a_{m_k} c_{i,m_k} \geq \sum_{j} a_{m_{s_j}} c_{i,m_{s_j}} = \sum_j c_{i,m_{s_j}} / c_{t(s_j),m_{s_j}} = \sum_j 1 = \infty. \end{align}

A counterexample to your specific question is obtained with $b_{j,k} = k^s (\log k)^j$, $c_{j,k} = k^{s + 1/(j+1)}$.

Answer 2

I don't think so. For your sequence $(a_k)$ let $(c_k)$ be sequence $(a_k/f(k))$, where $f(k)$ grows faster than $\log(k)^b$ for any $b>0$ but slower than $k^\epsilon$ for any $\epsilon>0$. It's straightforward to show that such $f$ exists.

Now, $\Sigma a_kk^s<\infty$ implies $\Sigma c_kk^s<\infty$ and $\Sigma a_kk^{s+\epsilon}=\infty$ implies $\Sigma c_kk^{s+\epsilon}=\infty$. Finally, $\Sigma c_kk^s(\log k)^b<\infty$, which contradicts your proposition.

Answer 3

This seems false. As a counterexample, let $$a_n=\frac{1}{n\cosh{(\sqrt{\ln{n}})}}$$ and let $s=0.$