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$\gamma$ is the lower incomplete gamma function. Is $\gamma(1, x) \ge \gamma(k, kx)$ when $k \in Z^+$, $x \in (0,1)$?

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Have you tried anything? –  Galois Group Aug 10 '12 at 23:07
    
@Fortuon: I don't know if your question is serious, but I tried some algebra, some calculus, wolfram alpha, and googling for bounds. –  martin Aug 10 '12 at 23:14
    
Welcome to Math.Stackexchange! It is often helpful if you list/describe what you have already done, so that potential answerers can see the dead ends you have been in and avoid them. –  Galois Group Aug 10 '12 at 23:17
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Have you seen this, by any chance? –  J. M. Aug 11 '12 at 0:42
    
@J.M. yes I saw that when I was "googling for bounds" but I was not able to convert it into a solution... –  martin Aug 11 '12 at 1:22

1 Answer 1

It doesn't appear to be the case.

Addendum: Notice that $\gamma(1,x) = 1-e^{-x}$, so $\gamma(1,x) < 1$ for $x\in(0,1)$. But $$\gamma(k,k x) \sim \frac{(k x e^{-x})^k}{k(1-x)} \qquad (k\to\infty).$$ (See DLMF 8.11.6.) Thus, for any $x\in(0,1)$ we can find a $k$ large enough so $\gamma(k,k x) > 1 > \gamma(1,x)$.

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