A special pair of sequences in $c_0$
This could be a rather trivial question but I've spent some time on it and I can't come up with anything:
Do there exist sequences of positive numbers, say $(\alpha_n),(\beta_n)$ in $c_0$, i.e. $\alpha_n\to0$ and $\beta_n\to0$ such that the following two conditions are satisfied:
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$\sum_{n=1}^\infty n\beta_n<\infty$
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there exists $\delta>0$ and a subsequence of indices $n_1<n_2<n_3<\dots$ such that $\beta_{n_k}\cdot\sum_{j=1}^{n_k}\alpha_j\ge\delta$, i.e. the sequence $\{\beta_n\sum_{j=1}^n\alpha_j\}_{n=1}^\infty$ does not converge to $0$.
All I have tried is to play around with some standard sequences, but once I get the one condition satisfied, the other one breaks down. I would appreciate any help!
A comment: in order for condition 2 to be satisfied, one needs to take $\alpha_n$ to be a sequence that is not in $\ell^1$. On the other hand, for condition (1) to be satisfied $(\beta_n)$ not only has to be in $\ell^1$, but it has to converge "fast enough".
Solution 1:
No, there do not exist such sequences. Since $\alpha_n=o(1)$, we must have $$\sum_{n=1}^N{\alpha_n}=\sum_{n=1}^N{o(1)}=o(N)$$
Thus $$\sum_{N=1}^{\infty}{\left(\beta_N\sum_{n=1}^N{\alpha_n}\right)}=\sum_{N=1}^{\infty}{o(N)\beta_N}\lesssim\sum_{N=1}^{\infty}{N\beta_N}<\infty$$ where ${\lesssim}$ indicates dropping a constant prefactor. Since $\sum_{N=1}^{\infty}{\left(\beta_N\sum_{n=1}^N{\alpha_n}\right)}$ is summable, its terms must tend to $0$, whence the claim.
Solution 2:
Ok, here is my attempt... my concern is that I've used a weak consequence of the convergence of the series $\sum n\beta_n$.
$n\beta_n \rightarrow 0$. So $\forall \epsilon>0$, $\exists N. n>N \Rightarrow n\beta_n < \epsilon$.
Also, $\forall \epsilon>0$, $\exists M . n>M \Rightarrow \alpha_j < \epsilon$.
Let $B = \max\{a_j: j \leq M$}.
If $n>\max(N,M)$, and $k>0$, $$ \beta_{n+k}\sum_{j=1}^{n+k}a_j \leq \beta_{n+k}(nB + k\epsilon) < \frac{\epsilon}{n+k}(nB + k\epsilon) $$
Pick $k$ such that $\frac{nB}{n+k} < \epsilon$. Then $$ \beta_{n+k}\sum_{j=1}^{n+k}a_j < 2\epsilon^2 $$
So $\beta_n\sum^n a_j$ converges to 0.