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.
Best Answer
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.