Integral Form of Gamma Function equivalent to Euler Form/Lemma

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Lemma for Integral Form of Gamma Function equivalent to Euler Form

Let $0 \le t \le m$.

Then:

$0 \le e^{-t} - \paren {1 - \dfrac t m}^m \le t^2 \dfrac {e^{-t} } m$


Proof

From Exponential of x not less than 1+x:

$1 + x \le e^x$

Let $x = \pm \dfrac t m$.


Then:

\(\ds 1 - \dfrac t m\) \(\le\) \(\ds e^{-t/m}\)
\(\ds \leadsto \ \ \) \(\ds \paren {1 - \dfrac t m}^m\) \(\le\) \(\ds e^{-t}\)

and:

\(\ds 1 + \dfrac t m\) \(\le\) \(\ds e^{t/m}\)
\(\ds \leadsto \ \ \) \(\ds \paren {1 + \dfrac t m}^m\) \(\le\) \(\ds e^t\)


Then:

\(\ds e^{-t}\) \(\ge\) \(\ds \paren {1 - \dfrac t m}^m\)
\(\ds \) \(=\) \(\ds e^t \paren {1 - \dfrac t m}^m e^{-t}\)
\(\ds \) \(\ge\) \(\ds e^{-t} \paren {1 - \dfrac t m}^m \paren {1 + \dfrac t m}^m\)
\(\ds \) \(=\) \(\ds e^{-t} \paren {1 - \dfrac {t^2} {m^2} }^m\) Difference of Two Squares
\(\ds \) \(\ge\) \(\ds e^{-t} \paren {1 - \dfrac {t^2} m}\) Corollary to Bernoulli's Inequality


Hence:

\(\ds 0\) \(\le\) \(\ds e^{-t} - \left({1 - \dfrac t m}\right)^m\)
\(\ds \) \(\le\) \(\ds t^2 \dfrac {e^{-t} } m\)

$\blacksquare$


Sources

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