No (Unless I made some mistake )

We claim the answer is no, it isn't always guranteed to diverge by constructing a converging sequence. All sequences worked on here are strictly positive. Also without loss of generality the logarithm is in base $2$ Lemma 1: Call a sequence of reals $\{ s_i \}_{i \geq 0}$ to be good if there's a strictly decreasing sequence $\{ x_i \}_{i \geq 1}$ of real numbers such that $s_k = 2^k x_{2^k}$. Then $\{s_i \}$ is good iff for every $j$, we have $s_j < 2 s_{j-1}$. ProofOne direction is trivial to see, if the sequence is good then $2s_{j-1} - s_j = 2^j x_{j-1} - 2^j x_j = 2^j (x_{j-1} - x_j) > 0$. For the other direction, assume $s_j < 2s_{j-1}$, and define $x_{2^k} = \frac{s_k}{2^k}$, and for $j \in [0, 2^k]$, $x_{2^k+j} = x_{2^{k+1}} + (x_{2^k} - x_{2^{k+1}}) \frac{j}{2^k}$. It's easy to see this sequence is monotonic. Lemma 2: We can find a sequence $\{s_i \}_{i \geq 1}$ such that $\{s_i \}$ is good, i.e $s_i < 2s_{i-1}$ for all $i$ $\sum_{i \leq 1} s_i$ is divergent $\sum_{i \leq 1} \min(s_i, \frac{1}{i})$ is convergent ProofWe will construct a sequnce $\{s_i \}$ and a monotonically increasing sequence of positive integers $\{a_i \}_{i \geq 0}$ with the following properties (it's easy to see the lemma follows if we are able to construct sequence with such property): For $n \in [a_{2i-1}, a_{2i})$, we have $s_n < \frac{1}{n}$, for $n \in (a_{2i}, a_{2i+1}]$, we have $\frac{1}{n} < s_n$ and $s_{a_{2i}} = \frac{1}{a_{2i}}$ $\sum_{j \in [a_{2i-1}, a_{2i})} s_j < \frac{1}{2^i}$ $\sum_{j \in [a_{2i}, a_{2i+1})} s_j > 1$ $\sum_{j \in [a_{2i}, a_{2i+1})} \frac{1}{j} \sim \frac{1}{2^i}$ (We say $b_i \sim c_i$ if $b_i = \Theta(c_i)$) Let $c = \frac{1.9}{1.9-1}$, so observe $1 + 1.9^{-1} + 1.9^{-2} + \cdots < c$. Also set $a_{-1} = 0$. To construct $a_{2i}, a_{2i+1}$, pick a really large $M_i$ such that $\frac{1}{cM_i} < \frac{1}{2^i}$, then pick an even larger $N_i > 1000M_i$ such that $\frac{1.9^{N_i(e^{\frac{1}{2^i}}-1)}}{N_i} > 1$ (we can find because top is exponential in $N_i$ and bottom is linear, so eventually top dominates the bottom). Now set $a_{2i} = N_i$, and $a_{2i+1} = e^{\frac{1}{2^i}}N_i$. Now define $s$ in $[a_{2i-1}+1, a_{2i+1}]$ as follows: $s_{a_{2i}} = \frac{1}{a_{2i}}$ For $a_{2i-1} < j \leq a_{2i}$ with $j = a_{2i} - k$, let $s_j = 1.9^{-j} s_{a_{2i}}$ For $a_{2i} < j \leq a_{2i+1}$ with $j = a_{2i} + k$ let $s_j = 1.9^j s_{a_{2i}}$. I claim this satisifies all the conditions. (1) is easy to see. (2) follows since $\sum_{j \in [a_{2i-1}, a_{2i}]} s_j < s_{a_{2i}} (1 + 1.9^{-1} + 1.9^{-2} + \cdots) = c s_{a_{2i}} < \frac{1}{2^i}$. Also (3) follows since $\sum_{j \in [a_{2i}, a_{2i+1}]} s_{j} > s_{a_{2i+1}} = 1.9^{a_{2j+1}-a_{2j}} s_{a_{2j}} = 1.9^{N_i(e^\frac{1}{2^i}-1)} \frac{1}{N_i} > 1$. (4) follows since $\sum_{j \in [a_2i, a_{2i+1}) } \frac{1}{j} \sim \log a_{2i+1} - \log a_{2i} = \log (\frac{a_{2i+1}}{a_{2i}}) = \log (e^{\frac{1}{2^i}}) = \frac{1}{2^i}$, hence done. Lemma 3 If $\{x_i \}$ and $\{t_i \}$ is monotonically decreasing, then $\{ \min(x_i, t_i) \}$ is monotonically decreasing too. ProofWe have $x_i > x_{i+1}$, $t_i > t_{i+1}$. If $\min(x_i, t_i), \min(x_{i+1}, t_{i+1})$ both come from the same sequence then we're done, otherwise if $\min(x_{i+1}, t_{i+1}) = x_{i+1}$ and $\min(x_i, t_i) = t_i$, then $\min(x_{i+1}, t_{i+1}) = x_{i+1} \leq t_{i+1} < t_i = \min(x_i, t_i) $ Returning to the main problem, let $\{s_i \}$ be the sequence constructed in Lemma-2. By Lemma-1, we can find a monotonically decreasing $\{x_i \}$ such that $s_i = 2^i x_{2^i}$. I claim that this sequence $\{x_i \}$ satisfies the hypothesis of the problem. By the Cauchy condensation test applied on $\sum x_i$, it diverges because $\sum s_i$ diverges. Define $t_i = \min(x_i, \frac{1}{i \log i})$. By Lemma-3, $t_i$ is strictly decreasing. Also observe $2^k t_{2^k} = \min(s_k, \frac{1}{k})$. So again by Cauchy condensation theorem, $\sum t_k = \sum \min(x_i, \frac{1}{i \log i})$ converges because $\sum 2^k t_{2^k} = \sum \min(s_i, \frac{1}{i})$ converges.