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  "body": "\n## Proof\n\nWe express $E[X]$ as a series, evaluate it by exchanging a double sum, and then pass to $T = X + 1$.\n\n1. Let $p \\in (0, 1]$, $q = 1 - p \\in [0, 1)$, $S = \\set{0, 1, 2, \\dots}$ and $p(k) = q^k p$ for $k \\in S$.\n2. Let $\\mu$ be the [[law|0TST4MHA3A]] of $X$. Since $X$ has the [[geometric distribution|0XCVV5Q500]], $\\mu(B) = \\sum_{k \\in B \\cap S} p(k)$ for each Borel set $B \\subseteq \\R$. In particular $\\mu(\\set{k}) = p(k)$ for $k \\in S$.\n3. Let $h \\colon \\R \\to [0, \\infty)$ be Borel measurable. We show $\\int_\\R h \\, d\\mu = \\sum_{k \\in S} h(k) \\, p(k)$ in $[0, \\infty]$, using the [[integral of a non-negative function|GEA8H3228S]].\n4. Let $s = \\sum_{i=1}^{m} a_i \\mathbf{1}_{A_i}$ be a [[simple function|7WMBX7M4P6]] with $0 \\le s \\le h$, $a_i \\ge 0$ and pairwise disjoint Borel sets $A_i$. By step 2, $\\int s \\, d\\mu = \\sum_{i} a_i \\sum_{k \\in A_i \\cap S} p(k)$.\n5. Each $k \\in S$ lies in at most one $A_i$, and $s(k) = a_i$ when $k \\in A_i$. All terms are nonnegative, so the sum in step 4 equals $\\sum_{k \\in S} s(k) \\, p(k) \\le \\sum_{k \\in S} h(k) \\, p(k)$.\n6. Taking the supremum over all such $s$ gives $\\int h \\, d\\mu \\le \\sum_{k \\in S} h(k) \\, p(k)$.\n7. For each $N$, the function $s_N = \\sum_{k=0}^{N} h(k) \\mathbf{1}_{\\set{k}}$ is simple with $0 \\le s_N \\le h$. So $\\sum_{k=0}^{N} h(k) \\, p(k) = \\int s_N \\, d\\mu \\le \\int h \\, d\\mu$.\n8. Letting $N \\to \\infty$ in step 7 gives $\\sum_{k \\in S} h(k) \\, p(k) \\le \\int h \\, d\\mu$. With step 6, equality holds.\n9. We compute $\\sum_{k \\in S} k \\, p(k)$. For integers $n, k \\ge 0$ let $a_{n,k} = q^k p$ if $1 \\le n \\le k$, and $a_{n,k} = 0$ otherwise. All $a_{n,k}$ are nonnegative.\n10. For fixed $k$, we have $a_{n,k} = q^k p$ for the $k$ indices $1 \\le n \\le k$ and $a_{n,k} = 0$ otherwise. So $\\sum_{n=0}^{\\infty} a_{n,k} = k q^k p$.\n11. For fixed $n \\ge 1$ and $N \\ge n$, the [[finite geometric sum|05APWSX53S]] with ratio $q \\ne 1$ gives $\\sum_{k=n}^{N} q^k p = q^n p \\, \\dfrac{1 - q^{N-n+1}}{1 - q} = q^n (1 - q^{N-n+1})$.\n12. Since $0 \\le q \u003c 1$, we have [[$q^m \\to 0$|0Q7K4AXE53]]. So by step 11, $\\sum_{k=0}^{\\infty} a_{n,k} = q^n$ for $n \\ge 1$. For $n = 0$ this sum is $0$.\n13. By [[Fubini's theorem for series|WW53KG71KH]], both iterated sums of $(a_{n,k})$ equal the sum over all pairs. This sum does not depend on the order of the indices, because it is a supremum of finite sums.\n14. So steps 10 and 12 give $\\sum_{k=0}^{\\infty} k q^k p = \\sum_{n=1}^{\\infty} q^n$.\n15. By the finite geometric sum, $\\sum_{n=1}^{M} q^n = q \\, \\dfrac{1 - q^M}{1 - q}$. This tends to $\\dfrac{q}{1 - q} = \\dfrac{1 - p}{p}$ as $M \\to \\infty$, since $q^M \\to 0$.\n16. Step 3 with $h(x) = \\abs{x}$ and step 15 give $\\int \\abs{x} \\, \\mu(dx) = (1 - p)/p \u003c \\infty$.\n17. By the [[law of the unconscious statistician|01ZWX875AT]] with $g(x) = x$, $X$ is integrable and $E[X] = \\int_\\R x \\, \\mu(dx)$.\n18. By the definition of the [[Lebesgue integral|XVEWR93EMV]], $\\int x \\, \\mu(dx) = \\int x^+ \\, \\mu(dx) - \\int x^- \\, \\mu(dx)$, with $x^+ = \\max(x, 0)$ and $x^- = \\max(-x, 0)$.\n19. On $S$ we have $x^+ = x$ and $x^- = 0$. Step 3 and step 15 give $\\int x^+ \\, d\\mu = (1 - p)/p$ and $\\int x^- \\, d\\mu = 0$. So $E[X] = (1 - p)/p$.\n20. The constant random variable $1 = \\mathbf{1}_\\Omega$ is simple, with $E[1] = P(\\Omega) = 1$. So it is integrable.\n21. By [[linearity of expectation|0P7PTC8DDE]], $T = X + 1$ is integrable and $E[T] = E[X] + 1 = \\dfrac{1 - p}{p} + 1 = \\dfrac{1}{p}$.\n\n## Idea\n\nThe weight $k$ in $\\sum_k k q^k p$ is a count, $k = \\sum_{n=1}^{k} 1$. Writing it this way and summing in the other order turns the series into $\\sum_{n \\ge 1} P(X \\ge n)$, where each tail $P(X \\ge n) = q^n$ is a geometric sum. A last geometric series gives $(1 - p)/p$, and adding $1$ gives the mean of $T$.\n",
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