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:

*

*$\sum_{n=1}^\infty n\beta_n<\infty$


*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".
 A: 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.
A: 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.
A: The answer by Jacob Manaker is definitely the simplest approach, but I was convinced that there is a compact-operator argument that can disprove this existence. Here is what I came up with:
Exercise 1: A diagonal operator $x\in\mathbb{B}(\ell^2)$ (i.e. $x\xi_n=\alpha_n\xi_n$ for some bounded sequence $(\alpha_n)$ of complex numbers) is compact if and only if the diagonal sequence $(\alpha_n)$ belongs to $c_0$.
Exercise 2: An infinite matrix $[\alpha_{i,j}]_{i,j=1}^\infty$ satisfying $\sum_{i,j}|\alpha_{i,j}|^2<\infty$ defines a bounded operator on $\ell^2$ (by $\xi_n\mapsto[\alpha_{i,j}]\cdot\xi_n$, where $\xi_n$ is regarded as an infinite column vector with $0$ everywhere except the $n$ slot where the entry is $1$. Actually, one can go further and show that this condition forces the operator to be compact, but this we will not need.
Let $(\alpha_n),(\beta_n)\in c_0$ be positive sequences in $c_0$ with $\sum_nn\beta_n<\infty$. Consider $x$ to be the diagonal operator with $x\xi_n=\alpha_n$. By exercise 1 this is compact. Now it is well-known that compact operators form a two-sided ideal in $\mathbb{B}(\ell^2)$, so $TxT^*$ is also a compact operator for any $T\in\mathbb{B}(\ell^2)$. Set
$$T\equiv\begin{pmatrix}\beta_1^{1/2}&0&0&0&0&0&0&0&0&0&0&0&\dots\\0&\beta_2^{1/2}&\beta_2^{1/2}&0&0&0&0&0&0&0&0&0&\dots\\0&0&0&\beta_3^{1/2}&\beta_3^{1/2}&\beta_3^{1/2}&0&0&0&0&0&0&\dots\\0&0&0&0&0&0&\beta_4^{1/2}&\beta_4^{1/2}&\beta_4^{1/2}&\beta_4^{1/2}&0&0&\dots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}:=[\gamma_{i,j}]$$
Since $\sum_{i,j}|\gamma_{i,j}|^2=\sum_nn\cdot\beta_n<\infty$, $T$ defines a bounded operator over $\ell^2$. But with direct computations one sees that, for $n\neq m$ we have $\langle TxT^*\xi_n,\xi_m\rangle=0$, so $TxT^*$ is a diagonal operator. It is also a compact operator since the compacts are a two-sideed ideal. But by exercise 1 compact diagonal operators have a diagonal sequence that belongs to $c_0$. But computing $\langle TxT^*\xi_n,\xi_n\rangle$, which is precisely the diagonal sequence, one sees that
$$\langle TxT^*\xi_n,\xi_n\rangle=\beta_n\cdot\sum_{j=1}^n\alpha_j.$$ This proves that $\{\beta_n\sum_{j=1}^n\alpha_j\}\in c_0$, i.e. your sequence converges to $0$ and thus condition $2$ fails.
