using $\Gamma(z+1)=z\Gamma(z)$ and $\Gamma(z)=\lim\limits_{n\to+\infty}\frac{n!n^z}{z(z+1)\cdots(z+n)}$ proof that $$\psi(z+1)=-\lim_{n\to\infty}\left(\sum_{m=1}^{n}\frac{1}{m}-\ln n\right)+\sum_{l=1}^{\infty}\frac{z}{l(z+l)}$$
my attempt:
$$\begin{align} \Gamma(z+1)&=z\Gamma(z)\\ \ln\Gamma(z+1)&=\ln z\Gamma(z)\\ &=\ln z+\ln\Gamma(z)\\ &=\ln z+\ln\lim\limits_{n\to+\infty}\frac{n!n^z}{z(z+1)\cdots(z+n)}\\ &=\lim\limits_{n\to+\infty}\ln z+\ln\frac{n!n^z}{z(z+1)\cdots(z+n)}\\ &=\lim\limits_{n\to+\infty}\ln(n!n^z)-\sum_{l=1}^{n}\ln(z+l) \end{align}$$ but i dont know how to finish, did im done alright? there any hint?
Starting with the limit definition of the Gamma function, we have
$$\Gamma(z)=\lim_{n\to \infty}\frac{n!n^z}{\prod_{k=0}^n (z+k)}$$
Then, we have
$$\begin{align} \log \Gamma (z+1)&=\log z+\log \Gamma(z)\\\\ &=\log z+\lim_{n\to \infty}\left(\log(n!n^z)-\sum_{k=0}^n \log(z+k)\right) \end{align}$$
Now, the Digamma function can be written
$$\begin{align} \psi(z+1)&=\frac1z+\lim_{n\to \infty}\left(\log(n)-\sum_{k=0}^n \frac1{(z+k)}\right)\\\\ &=\lim_{n\to \infty}\left(\log(n)-\sum_{k=1}^n \frac1{(z+k)}\right)\\\\ &=\lim_{n\to \infty}\left(\log(n)-\sum_{k=1}^n \left(\frac1k-\frac z{k(z+k)}\right)\right)\\\\ &=-\lim_{n\to \infty}\left(\sum_{k=1}^n \frac1k -\log(n)\right)+\sum_{\ell=1}^{\infty}\frac{z}{\ell(z+\ell)} \end{align}$$