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  "body": "\n## Proof\n\nWe show that the partial sums form a Cauchy sequence, and then bound the norm of the limit.\n\n1. Let $t_N = \\sum_{n=1}^{N} \\|x_n\\|$ and $L = \\sum_{n \\ge 1} \\|x_n\\| = \\lim_N t_N$. The terms are nonnegative, so $t_N$ increases and $t_N \\le L$.\n2. For $M \u003e N$, $s_M - s_N = \\sum_{n=N+1}^{M} x_n$. By the [[triangle inequality|0FJZMBX0ZZ]] and induction on the number of terms, $\\|s_M - s_N\\| \\le \\sum_{n=N+1}^{M} \\|x_n\\| = t_M - t_N$.\n3. The real sequence $(t_N)$ converges, so it is a Cauchy sequence by [[this result|0QY87R9CY3]]. Given $\\varepsilon \u003e 0$, there is $N_0$ with $|t_M - t_N| \u003c \\varepsilon$ for $M, N \\ge N_0$.\n4. By step 2, $\\|s_M - s_N\\| \u003c \\varepsilon$ for $M, N \\ge N_0$. So $(s_N)$ is a [[Cauchy sequence|0MCK8K2D7Z]] in $V$.\n5. $V$ is a [[Banach space|0BDTTM24DP]], so it is [[complete|08EWT7J0SC]]. Hence $(s_N)$ [[converges|06JYPJ4WHH]] to some $s \\in V$.\n6. For each $N$, the triangle inequality gives $\\|s_N\\| \\le t_N \\le L$.\n7. Then $\\|s\\| \\le \\|s - s_N\\| + \\|s_N\\| \\le \\|s - s_N\\| + L$ for each $N$.\n8. Suppose $\\|s\\| \u003e L$. Let $\\varepsilon = \\|s\\| - L$. For large $N$, $\\|s - s_N\\| \u003c \\varepsilon$, and step 7 gives $\\|s\\| \u003c \\|s\\|$, a contradiction.\n9. Thus $\\|s\\| \\le \\sum_{n \\ge 1} \\|x_n\\|$.\n\n## Idea\n\nThe triangle inequality bounds the size of a block of terms of the series by the corresponding block of the convergent real series of norms. So the partial sums are Cauchy, and completeness gives a limit. The same inequality, passed to the limit, bounds the norm of the sum.\n",
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