According to Cauchy condensation test: for a non-negative, non-increasing sequence $(u_n)_{n \in \mathbb N}$ of real numbers, the series $\sum_{n \in \mathbb N} u_n$ converges if and only if the condensed series $\sum_{n \in \mathbb N} 2^n u_{2^n}$ converges.

The test doesn’t hold for any non-negative sequence. Let’s have a look at counterexamples.

A sequence such that $\sum_{n \in \mathbb N} u_n$ converges and $\sum_{n \in \mathbb N} 2^n u_{2^n}$ diverges

Consider the sequence $$u_n=\begin{cases} \frac{1}{n} & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ 0 & \text{ else} \end{cases}$$ For $n \in \mathbb N$ we have $$0 \le \sum_{k = 1}^n u_k \le \sum_{k = 1}^{2^n} u_k = \sum_{k = 1}^{n} \frac{1}{2^k} < 1,$$ therefore $\sum_{n \in \mathbb N} u_n$ converges as its partial sums are positive and bounded above. However $$\sum_{k=1}^n 2^k u_{2^k} = \sum_{k=1}^n 1 = n,$$ so $\sum_{n \in \mathbb N} 2^n u_{2^n}$ diverges.

A sequence such that $\sum_{n \in \mathbb N} v_n$ diverges and $\sum_{n \in \mathbb N} 2^n v_{2^n}$ converges

Consider the sequence $$v_n=\begin{cases} 0 & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ \frac{1}{n} & \text{ else} \end{cases}$$ We have $$\sum_{k = 1}^{2^n} v_k = \sum_{k = 1}^{2^n} \frac{1}{k} – \sum_{k = 1}^{n} \frac{1}{2^k} > \sum_{k = 1}^{2^n} \frac{1}{k} -1$$ which proves that the series $\sum_{n \in \mathbb N} v_n$ diverges as the harmonic series is divergent. However for $n \in \mathbb N$, $2^n v_{2^n} = 0$ and $\sum_{n \in \mathbb N} 2^n v_{2^n}$ converges.