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  "body": "\n## Proof\n\nWe count Dyck words as all balanced words minus the bad ones, and count the bad ones by reflection.\n\n1. Let $n$ be a natural number. By [[Dyck words are counted by Catalan numbers|017SRZ7NE9]], the [[Catalan number|092DRBFKMG]] $C_n$ is the number of [[Dyck words|0R9JR0465V]] of semilength $n$.\n2. If $n = 0$, then $C_0 = 1$, and $\\frac{1}{1} \\binom{0}{0} = 1 = \\frac{0!}{1!\\, 0!}$. From now on let $n \\ge 1$.\n3. For a word $p$ over $\\set{U, D}$, let $h(p)$ be its number of letters $U$ minus its number of letters $D$.\n4. Let $A$ be the set of words with $n$ letters $U$ and $n$ letters $D$. Such a word is fixed by the set of positions of its letters $U$, an $n$-element subset of $\\set{1, \\dots, 2n}$.\n5. So by the [[number of $k$-element subsets|0BFMR2FQ8B]], $|A|$ is the [[binomial coefficient|092GEF38YD]] $\\binom{2n}{n}$. Similarly, the set $A'$ of words with $n - 1$ letters $U$ and $n + 1$ letters $D$ has $|A'| = \\binom{2n}{n-1}$.\n6. A word of $A$ is a Dyck word unless some prefix $p$ has $h(p) \u003c 0$. Call such words bad, and let $A_{\\mathrm{bad}}$ be their set.\n7. Let $w \\in A_{\\mathrm{bad}}$, and let $w_{\\le t}$ be its prefix of length $t$. Each letter changes $h$ by $\\pm 1$, and $h$ of the empty prefix is $0$. Take the least $t$ with $h(w_{\\le t}) \u003c 0$. Then $h(w_{\\le t-1}) \\ge 0$, so $h(w_{\\le t}) = -1$.\n8. Define $\\rho(w)$ by exchanging $U$ and $D$ in all letters after position $t$. The suffix after $t$ has $h$ equal to $h(w) - h(w_{\\le t}) = 1$, and the exchange turns this into $-1$.\n9. So $h(\\rho(w)) = -1 - 1 = -2$. Since $\\rho(w)$ has length $2n$, it has $n - 1$ letters $U$ and $n + 1$ letters $D$. Thus $\\rho(w) \\in A'$.\n10. Conversely, let $v \\in A'$. Then $h(v) = -2 \u003c 0$, so as in step 7 the least $t$ with $h(v_{\\le t}) \u003c 0$ has $h(v_{\\le t}) = -1$. Exchanging $U$ and $D$ after position $t$ gives a word with $h = -1 + 1 = 0$, which is in $A_{\\mathrm{bad}}$.\n11. Both operations keep the first $t$ letters, and $t$ is defined from these letters alone. So they find the same $t$. Hence they are inverse to each other.\n12. By the [[bijection principle|0GV2573FP4]], $|A_{\\mathrm{bad}}| = |A'| = \\binom{2n}{n-1}$.\n13. By the [[complement principle|03QC59M99B]], $C_n = |A| - |A_{\\mathrm{bad}}| = \\binom{2n}{n} - \\binom{2n}{n-1}$.\n14. By the [[factorial formula|033KTKKRV6]], $\\binom{2n}{n} = \\frac{(2n)!}{n!\\, n!}$ and $\\binom{2n}{n-1} = \\frac{(2n)!}{(n-1)!\\, (n+1)!}$.\n15. By the definition of the [[factorial|0WFCGVMRRC]], $n! = n \\cdot (n-1)!$ and $(n+1)! = (n+1) \\cdot n!$. So $\\binom{2n}{n-1} = \\frac{n}{n+1} \\cdot \\frac{(2n)!}{n!\\, n!} = \\frac{n}{n+1} \\binom{2n}{n}$.\n16. By steps 13 and 15, $C_n = \\left(1 - \\frac{n}{n+1}\\right) \\binom{2n}{n} = \\frac{1}{n+1} \\binom{2n}{n}$.\n17. By step 14 and $(n+1) \\cdot n! = (n+1)!$, this equals $\\frac{(2n)!}{(n+1)!\\, n!}$.\n\n## Idea\n\nAmong the $\\binom{2n}{n}$ words with $n$ letters of each kind, the bad ones first reach level $-1$ somewhere. Flipping every letter after that point is a bijection onto all words with $n - 1$ letters $U$, so there are $\\binom{2n}{n-1}$ bad words. The difference simplifies to $\\frac{1}{n+1}\\binom{2n}{n}$.\n",
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