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  "body": "\n## Proof\n\n1. Let $A_N = \\sum_{n=0}^{N} a_n$ and $B_N = \\sum_{n=0}^{N} b_n$ be the [[partial sums|0NB643K60Q]]. Since $a_n \\le b_n$ for every $n$, we have $A_N \\le B_N$ for every $N$.\n2. Assume that $\\sum b_n$ converges, with sum $B$. Since $b_n \\ge 0$, the [[criterion for nonnegative series|0MX1HZBBXJ]] gives $B = \\sup_N B_N$. So $B_N \\le B$ for all $N$.\n3. By steps 1 and 2, $A_N \\le B$ for all $N$. So the partial sums of $\\sum a_n$ are bounded above.\n4. Since $a_n \\ge 0$, the same criterion shows that $\\sum a_n$ converges, with sum $\\sup_N A_N$.\n5. By step 3, $B$ is an upper bound of the $A_N$. So $\\sum a_n = \\sup_N A_N \\le B = \\sum b_n$.\n6. Now assume that $\\sum a_n$ diverges. If $\\sum b_n$ converged, then $\\sum a_n$ would converge by steps 2 to 4. So $\\sum b_n$ diverges.\n\n## Idea\n\nThe partial sums of the smaller series stay below those of the larger one. For nonnegative terms, convergence means bounded partial sums, so a bound for the larger series is also a bound for the smaller one.\n",
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