Approximation of $\Gamma(n+\lambda)/\Gamma(n)$ with $0<\lambda<1$ for large $n$ I can empirically show that it goes like $n^{\lambda}$ plus terms that go to $0$ when $n$ goes to infinity, but I was wondering if someone can actually provide an expression and roughly show me how to get there.
 A: It's easier to consider the reciprocal of your function, so define
$$ F(n):= \frac{\Gamma(n)}{\Gamma(n+\lambda)} = \frac{B(n,\lambda)}{\Gamma(\lambda)}. $$
The Beta-function can be given by the formula
$$ B(a,b) = \int_0^{1} x^{a-1}(1-x)^{b-1} \, dx, $$
so we can write 
$$ F(n) = \frac{1}{\Gamma(\lambda)}\int_0^{1} x^{n-1}(1+x)^{\lambda-1} \, dx. $$
If we now put $x=e^{-t}$, this becomes
$$ F(n) = \frac{1}{\Gamma(\lambda)}\int_0^{\infty} e^{-nt}(1-e^{-t})^{\lambda-1} \, dt, $$
to which Watson's lemma may be applied, which gives
$$ F(n) = n^{-\lambda} \left( 1 - \frac{\lambda(\lambda-1)}{2n} + O(n^{-2}) \right). $$
Hence
$$ \frac{1}{F(n)} = n^{\lambda} \left( 1 + \frac{\lambda(\lambda-1)}{2n} + O(n^{-2}) \right). $$
A: Use the Stirling's formula for the Gamma function:
$$\Gamma(t) =\sqrt{\frac{2 \pi}{t}} \left(\frac{t}{e} \right)^{t}(1+O(1/t)).$$
Hence
$$\frac{\Gamma(n+\lambda)}{\Gamma(n)}\sim\sqrt{\frac{2 \pi}{n+\lambda}} \left(\frac{n+\lambda}{e} \right)^{n+\lambda}\cdot\sqrt{\frac{n}{2 \pi}} \left(\frac{e}{n} \right)^{n}\sim n^{\lambda}$$
where we used the fact that $(1+\lambda/n)^n\cdot e^{-\lambda}\to 1$.
A: Use the asymptotic expression https://dlmf.nist.gov/5.11.E7 (which follows from Stirling)
$$\Gamma(n+\lambda) \sim \sqrt{2\pi} e^{-n}n^{n+\lambda-1/2}$$
then
$$\frac{\Gamma(n+\lambda)}{\Gamma(n)} \sim \frac{n^{n+\lambda-1/2}}{n^{n-1/2}} =
n^\lambda
$$
