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  "body": "\n## Proof\n\nWe argue by induction on $n$.\n\n1. For $n \\ge 0$ let $S_n = u_0 + \\sum_{k=0}^{n-1} b_k$ and $E_n = \\exp\\left( \\sum_{k=0}^{n-1} c_k \\right)$. The claim is $u_n \\le S_n E_n$.\n2. Since $u_0 \\ge 0$ and each $b_k \\ge 0$, we have $S_n \\ge 0$ and $S_{n+1} = S_n + b_n$.\n3. Since each $c_k \\ge 0$, the exponent of $E_n$ is non-negative. So $E_n \\ge 1$, because [[the exponential function is increasing|0KN2V7840E]]. Also $E_{n+1} = e^{c_n} E_n$.\n4. Base case: $S_0 = u_0$ and $E_0 = e^0 = 1$, so $u_0 \\le S_0 E_0$ holds with equality.\n5. Induction step: assume $u_n \\le S_n E_n$ for some $n \\ge 0$.\n6. Since $1 + c_n \\ge 1 \u003e 0$, multiplying gives $(1 + c_n) u_n \\le (1 + c_n) S_n E_n$.\n7. By [[the inequality 1 + x ≤ exp x|012C7GFF20]], $1 + c_n \\le e^{c_n}$. Since $S_n E_n \\ge 0$ by steps 2 and 3, $(1 + c_n) S_n E_n \\le e^{c_n} S_n E_n = S_n E_{n+1}$.\n8. Since $b_n \\ge 0$ and $E_{n+1} \\ge 1$, we have $b_n \\le b_n E_{n+1}$.\n9. By the hypothesis and steps 6 to 8, $u_{n+1} \\le (1 + c_n) u_n + b_n \\le S_n E_{n+1} + b_n E_{n+1} = S_{n+1} E_{n+1}$.\n10. By induction, $u_n \\le S_n E_n$ for each $n \\ge 0$. This is the stated inequality.\n\n## Idea\n\nAt each step the bound is multiplied by at most $1 + c_k \\le e^{c_k}$, and the new term $b_k$ is added. Since the exponential factor is at least $1$, the added terms can be put inside the same factor, so the whole bound is a sum times one exponential.\n",
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