If $0\le a_k\le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$, and divergence of $\sum a_k$ gives divergence of $\sum b_k$

Let $(a_k)$ and $(b_k)$ be sequences of reals and suppose there is $K\in \mathbb{N}$ with

$0\le a_k\le b_k\text{for all }k\ge K\mathrm{.}$

Then:

1. if $\sum b_k$ converges then $\sum a_k$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series);
2. if $\sum a_k$ diverges then $\sum b_k$ diverges.

The same statement holds verbatim for series with a general starting index $m$, applied to the shifted sequences of Series, partial sums, convergence and the sum, divergence, and the tail series.

The hypothesis is on the terms from some index on, not on all of them: finitely many terms of either sequence may violate it, or be negative, without affecting the conclusion. What may not be dropped is nonnegativity of $(a_k)$ from that index on.

• Both nonnegativity hypotheses are used, and in different places. $a_k\ge 0$ is what lets convergence of ${\sum }_{k\ge K}{a}_k$ be read off from boundedness of its partial sums, and $b_k\ge a_k\ge 0$ is what makes the sum of ${\sum }_{k\ge K}{b}_k$ an upper bound for the partial sums ${\beta }_j$. Drop the sign hypothesis and the theorem is false, not merely unproved; the companion page exhibits a pair with $a_k\le b_k$ for every $k$, $\sum b_k$ convergent and $\sum a_k$ divergent.

• The comparison is with a series, not with a limit. No quotient $a_k\mathrm{/}{b}_k$ appears and no $b_k$ is required to be nonzero, which is what distinguishes this test from the limit comparison test proved next.