Cauchy condensation test

Statement

Let $(a_n)_{n \ge 1}$ be a decreasing real sequence with $a_n \ge 0$ for all $n$. Then $\sum_{n=1}^{\infty} a_n$ converges if and only if the condensed series $\sum_{k=0}^{\infty} 2^k a_{2^k}$ converges.

Remarks

The proof groups the terms from $2^k$ to $2^{k+1} - 1$. Since the sequence decreases, $\sum_{n=1}^{\infty} a_n \le \sum_{k=0}^{\infty} 2^k a_{2^k} \le 2 \sum_{n=1}^{\infty} a_n$. With $a_n = 1/n^p$ it gives the p-series test. With $a_n = 1/(n \ln n)$ for $n \ge 2$, the condensed series is a multiple of $\sum 1/k$, so $\sum 1/(n \ln n)$ diverges.

Explanations

L1 human

For a decreasing sequence, the terms between $2^k$ and $2^{k+1}$ are all about the same size as $a_{2^k}$. So only one term out of each such block matters, weighted by the size $2^k$ of the block.