Bounding the Gamma Function I'm trying to verify a bound for the gamma function
$$ \Gamma(z) = \int_0^\infty e^{-t}t^{z - 1}\;dt. $$
In particular, for real $m \geq 1$, I'd like to show that
$$ \Gamma(m + 1) \leq 2\left(\frac{3m}{5}\right)^m. $$
Knowing that the bound should be attainable, my first instinct is to split the integral as
$$ \Gamma(m + 1) = \int_0^{3m/5} e^{-t}t^{m}\;dt + \int_{3m/5}^\infty e^{-t}t^m\;dt
\leq (1 - e^{-3m/5})\left(\frac{3m}{5}\right)^m + \int_{3m/5}^\infty e^{-t}t^m\;dt. $$
Using integration by parts,
$$ \int_{3m/5}^\infty e^{-t}t^m\;dt = e^{-3m/5}\left(\frac{3m}{5}\right)^m + m\int_{3m/5}^\infty e^{-t}t^{m-1}\;dt.$$
So the problem has been reduced to showing
$$ m\int_{3m/5}^\infty e^{-t}t^{m-1}\;dt \leq \left(\frac{3m}{5}\right)^m. $$
But this doesn't seem to have made the problem any easier.
Any help is appreciated, thanks.
 A: Suppose $m = n + \alpha$ where $0 \le \alpha < 1$, so that $\alpha$ is the fractional part of $m$. Taking logarithms, the inequality becomes
$$ \sum_{k=1}^n \log (k + \alpha) + \log \Gamma (1+\alpha) <
\log 2 + (n+\alpha) \log (n+\alpha) + (n+\alpha) \log \frac{3}{5} $$
Now using Riemann sums, we have
$$ \sum_{k=1}^n \log (k + \alpha) < \int_1^{n+1} \log(x+\alpha) dx$$
which is
$$ (n+1+\alpha) \log (n+1+\alpha) - n - 1 - (1+\alpha) \log(1+\alpha) + 1.$$
This means we need to show that
$$ (n+1+\alpha) \log (n+1+\alpha) - n - (1+\alpha) \log(1+\alpha) + \log \Gamma (1+\alpha) < \log 2 + (n+\alpha) \log (n+\alpha) + (n+\alpha) \log \frac{3}{5} $$
Rearranging terms, this is
$$ (n+1+\alpha) \log (n+1+\alpha) -  (n+\alpha) \log (n+\alpha) <
\log 2 - \log \Gamma (1+\alpha) + (n+\alpha) \log \frac{3}{5} + n + (1+\alpha) \log(1+\alpha)$$
Now for the LHS we have
$$ \log (n+1+\alpha) + \log \left( 1 + \frac{1}{n+\alpha}\right)^{n+\alpha} <
\log (n+1+\alpha) + 1$$
because $ \log \left( 1 + \frac{1}{x}\right)^x < 1$ by a trivial calculation.
On the RHS we have
$$ n \log\frac{3e}{5} + \log 2 - \log \Gamma (1+\alpha) + \alpha \log \frac{3}{5}  + (1+\alpha) \log(1+\alpha) > \log (n+1+\alpha) + 1$$
for all $n > n_0$ for some $n_0$ because the coefficient $\log\frac{3e}{5}$ on $n$ is positive and the remaining terms are bounded by a constant. This concludes the proof. Note that the proof also goes through with a factor of $\frac{2}{5}$, just barely, and requiring $n_0 = 22.$ The original post has $n_0 = 1.$
I'm not sure of all the details but I hope it's a start.
A: I'll prove something that's close enough for my applications; in particular, that $$\Gamma(m + 1) \leq 3\left(\frac{3m}{5}\right)^m.$$
Let $0 < \alpha < 1$ be chosen later.  We'll split $e^{-t}t^m$ as $(e^{-\alpha t}t^m)e^{-(1 - \alpha)t}$ and use this to bound the integral.
First, take a derivative to find a maximum for $e^{-\alpha t}t^m$.
$$\frac{d}{dt}e^{-\alpha t}t^m = -\alpha e^{-\alpha t}t^m + me^{-\alpha t}t^{m-1} = -\alpha e^{-\alpha t}t^{m - 1}\left(t - \frac{m}{\alpha}\right). $$
So $t = m / \alpha$ is a critical point, and in particular a maximum (increasing before and decreasing after, if you like).
Then we can bound the integral
$$ \Gamma(m + 1) = \int_0^\infty e^{-t}t^m\;dt \leq \left(\frac{m}{\alpha e}\right)^m \int_0^\infty e^{-(1 - \alpha)t}\;dt = \left(\frac{m}{\alpha e}\right)^m \left(\frac{1}{1 - \alpha}\right).$$
Choosing $\alpha = 5/(3e)$ and noting that $\frac{1}{1 - 5/(3e)} \leq 3$, we've proven
$$ \Gamma(m + 1) \leq 3\left(\frac{3m}{5}\right)^m. $$
