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Let $X$ be a geometric random variable i.e. it represents the number of consecutive failures before you get the first success where the success probability is $\rho$. We know $E[X] = 1/\rho$ and $E[X^2] = (1-\rho)/(\rho)^2$. Does it generally hold that $E[X^c] \leq O(1/\rho^c)$ ? If yes is there a short proof ?

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For every $k\geqslant1$, $$ E(X(X-1)\cdots(X-k+1))=\frac{k!\,(1-\varrho)^{k-1}}{\varrho^k}\leqslant\frac{k!}{\varrho^k}, $$ and $X^k$ is a linear combination of the random variables $X(X-1)\cdots(X-i+1)$ for $i\leqslant k$ hence, indeed, there exists absolute constants $C_k$ such that $$ E(X^k)\leqslant\frac{C_k}{\varrho^k}, $$ The most direct way to prove the first identity above might be to differentiate $k$ times at $s=1$ the identity $$ E(s^X)=\sum_{i=1}^\infty \varrho(1-\varrho)^{i-1}s^i=\frac{\varrho s}{1-(1-\varrho)s}=\frac{\varrho}{1-\varrho}\,\left(\frac1{1-(1-\varrho)s}-1\right). $$

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