Q. Use the definition $$ \Gamma(x) = \lim_{n\to \infty} \frac{n!n^x}{x(x+1)\cdots(x+n)} $$ in this problem.
a) Show that $ \Gamma (x+1) = x\Gamma(x) $ for $ x>0 $
b) First, express $\mathbb{ln}\Gamma(x)$ as sum of terms. Then, differentiate to show that $$\Gamma'(1) = -\gamma$$ where $\gamma$ is the Euler constant.
For part a), I tried putting $(x+1)$ into the expression given and tried to simplify, but I didn't get much far. Same for multiplying x to the definition as well. For part b), I am completely lost as I do not know how to express $\mathbb{ln}\Gamma(x)$ as a sum of terms.
a) Separate out the required terms for the expression for x $$ \frac{n!n^{x+1}}{(x+1)(x+1)\dots(x+1+n)} = \frac{n!n^x}{x(x+1)\dots(x+n)}\cdot\frac{nx}{x+n+1} $$ Both factors converge independently for $n\to\infty$.
b) The logarithmic derivative of the term under the limit is $$ \ln(n)-\frac1x-\frac1{x+1}-\dots-\frac1{x+n} $$ Compare with the definition of the Euler-Mascheroni constant.
Now argue that you can exchange the limits of the derivative and the gamma function. The logarithm is known as continuous.