Proof of an inequality under Equivalence of Weierstrass and Euler's Definitions of Gamma Function

I am using my lecturer's notes on Special Functions. When dealing with Gamma Functions under the title Equivalence of Wierstrass and Euler's Definitions, he has used a lemma (with no proof). I tried hard but couldn't get a way how to start.

Lemma: If $0 \le t <n$ /; $n \in \mathbb{Z}^+$, then this inequality holds: $$0 \le \exp{(-t)} -{\left( {1-\frac{t}{n}} \right)}^n \le \frac{t^2 \exp{(-t)}}{n}$$

• Not as easy as @Jonas Meyer answer, but here is my try: $$\exp{(t/n)} =1 + \frac{t}{n} + \sum_{i=2}^{\infty} \frac{t^i}{i! n^i}$$ and $${\left({1-\frac{t}{n}}\right)}^{-1}= 1+\frac{t}{n}+\sum_{i=2}^{\infty} {\left({\frac{t}{n}}\right)}^i$$ On Comparing, $${\left({1+\frac{t}{n}}\right)} \le \exp{(t/n)} \le {\left({1-\frac{t}{n}}\right)}^{-1}$$ Taking $n$th power of each term $${\left({1+\frac{t}{n}}\right)}^{n} \le \exp{(t)} \le {\left({1-\frac{t}{n}}\right)}^{-n}$$ or, $${\left({1+\frac{t}{n}}\right)}^{-n} \ge \exp{(-t)} \ge {\left({1-\frac{t}{n}}\right)}^{n}$$ Now? Out of Ideas. – gaurav Dec 29 '11 at 9:01
• Reached the limit of words. Above relation easily implied first inequality. – gaurav Dec 29 '11 at 9:05

Multiplying by $e^t$, this is equivalent to showing that $\displaystyle{0\leq 1-e^t\left(1-\frac{t}{n}\right)^n\leq \frac{t^2}{n}}$. Let $\displaystyle{f(t)=1-e^t\left(1-\frac{t}{n}\right)^n}$. Then $\displaystyle{f'(t)=\frac{t}{n}e^t\left(1-\frac{t}{n}\right)^{n-1}\geq 0}$ for all $t\in[0,n]$, and $f(0)=0$, so $f(t)\geq 0$ for all $t\in[0,n]$, proving the first inequality.
For the second inequality, first suppose $n\geq 2$. Consider $g(t)=\frac{t^2}{n}$, and note that by the racetrack principle it suffices to show that $f'(t)\leq g'(t)$ for all $t\in[0,n]$. This reduces to showing that $\displaystyle{e^t\left(1-\frac{t}{n}\right)^{n-1}\leq 2}$ for all $t\in[0,n]$. If we let $\displaystyle{h(t)=e^t\left(1-\frac{t}{n}\right)^{n-1}}$, then $\displaystyle{h'(t)=e^t\left(1-\frac{t}{n}\right)^{n-2}\cdot\frac{1-t}{n}}$, which shows that $h$ reaches its maximum at $t=1$. Thus the second inequality, in the case $n\geq 2$, is reduced to proving that $\displaystyle{e\left(1-\frac{1}{n}\right)^{n-1}\leq 2}$. You can check that the inequality holds when $n=2$, and you can show that $\displaystyle{\left(1-\frac{1}{n}\right)^{n-1}}$ decreases as $n$ increases.
This just leaves the $n=1$ case for the second inequality, or $1-e^t(1-t)\leq t^2$. Taking the derivative of $t^2-1+e^t(1-t)$ shows that this function increases on $(0,\ln 2)$ and decreases on $(\ln(2),1)$, and it is $0$ when $t=0$ or $t=1$. This implies that $t^2-1+e^t(1-t)\geq 0$ for all $t\in[0,1]$. Alternatively, when $t<1$, factoring out $1-t$ reduces the inequality to the well known fact that $e^t\geq 1+t$.
• Thanks, but it is not true that $${\left( {1-\frac{1}{n}}\right)}^{n-1}$$ decreases when $n$ increases. I think this is the easiest way to approach the result. – gaurav Dec 29 '11 at 8:32
• @gaurav: It is in fact true that $\displaystyle{\left( {1-\frac{1}{n}}\right)}^{n-1}$ decreases when $n$ increases. A straightforward way to show this is to define $\displaystyle{k(x)=\left( {1-\frac{1}{x}}\right)}^{x-1}$, and show that $k'(x)<0$ when $x>1$. – Jonas Meyer Dec 29 '11 at 8:37