Trying to prove that $\lim_{n\rightarrow\infty}(\frac{\Gamma '(n+1)}{n!} -\log(n))=0$ In my attempt to prove that $\Gamma'(1)=-\gamma$, I've reduced the problem to proving that $\lim_{n\rightarrow\infty}(\frac{\Gamma '(n+1)}{n!} -\log(n))=0$.
Where $\gamma$ is the Euler-Mascheroni constant, and $\log$ denotes the natural logarithm.
I've been messing with it for a while without achieving much of anything.  The first derivative of the Gamma function does have a recursive formula which can be found through iterated integration by parts, but that was what I used to get where I am, and applying it again just takes me back to where I started.
My book lists a ton of equivalent definitions for the Gamma function, but only gives the integral definition for its derivatives and I've just had considerable trouble doing much with that integral.
I should note that at first I was trying to prove that $\frac{\Gamma'(n+1)}{n!}\sim\log(n)$, but now I'm pretty sure that showing that their difference in the limit is zero would be sufficient, since what I'm ultimately interested in is showing that $\lim_{n\rightarrow\infty}(-\sum_{k=0}^n\frac{1}{k} +\frac{1}{n!}\Gamma'(n+1)) =-\gamma$
Hopefully someone can help me with this.  Thanks.
 A: We have $\log(\Gamma(n+2))-\log(\Gamma(n+1))=\log(n+1)$, so by the mean value theorem, $\frac{\Gamma'}{\Gamma}(s)=\log(n+1)$ for some $s\in[n+1,n+2]$.  Now, $\frac{\Gamma'}{\Gamma}$ is increasing, so repeating the argument on the interval $[n,n+1]$, we get $\log(n)\leq\frac{\Gamma'}{\Gamma}(n+1)\leq \log(n+1)$.  The result now follows, as $\lim_{n\to\infty}\log(n+1)-\log(n)=0$.
A: If what you want is to prove that $\Gamma'(1)=-\gamma$, I will show you a very slick solution I learned from user robjohn.
(I guess) you know that $$\int_0^1 \frac{1-x^n}{1-x}dx=H_n$$
(just expand the function it is easy) 
Now look at
$$I(n)=\int_0^1 x^{n}\log(1-x)dx$$
We integrate by parts, to get
$$I(n)=\int_0^1 x^{n}\log(1-x)dx=\left.\frac{1-x^{n+1}}{n+1}\log(1-x)\right|_0^1- \frac 1{n+1}\int_0^1 \frac{1-x^{n+1}}{1-x}dx$$
$$I(n)=\int_0^1 x^{n}\log(1-x)dx=-\frac 1{n+1}\int_0^1 \frac{1-x^{n+1}}{1-x}dx$$
$$I(n)=\int_0^1 x^{n}\log(1-x)dx=-\frac {H_{n+1}}{n+1}$$
Now let $x=1-u$, then $un=m$,
$$I(n)=\int_0^1 (1-u)^{n}\log(u)du$$
$$I(n)=\frac 1 n\int_0^n \left(1-\frac m n\right)^{n}\log\left(\frac m n\right)dm$$
$$I(n)=\frac 1 n\int_0^n \left(1-\frac m n\right)^{n}\log ( m )dm-\frac 1 n\int_0^n \left(1-\frac m n\right)^{n}\log( n)dm$$
Now we use the last equiality we derived, to start getting into something:
$$-\frac {n}{n+1}H_{n+1}=\int_0^n \left(1-\frac m n\right)^{n}\log ( m )dm-\log n \int_0^n \left(1-\frac m n\right)^{n}dm$$
Now the rightmost integral is just 
$$\int_0^n {{{\left( {1 - \frac{m}{n}} \right)}^n}} dm = n\int_0^1 {{{\left( {1 - u} \right)}^n}} du = n\int_0^1 {{u^n}} du = \frac{n}{{n + 1}}$$
$$-\frac{n}{{n + 1}}{H_{n + 1}} = \int_0^n {{{\left( {1 - \frac{m}{n}} \right)}^n}} \log mdm - \frac{n}{{n + 1}}\log n$$
So we get
$$\frac{n}{{n + 1}}\left( {\log n - {H_{n + 1}}} \right) = \int_0^n {{{\left( {1 - \frac{m}{n}} \right)}^n}} \log mdm$$
Now, by letting $n\to \infty$, we get
$$\eqalign{
  & \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}}\left( {\log n - {H_{n + 1}}} \right) = \mathop {\lim }\limits_{n \to \infty } \int_0^n {{{\left( {1 - \frac{m}{n}} \right)}^n}} \log mdm  \cr 
  &  - \gamma  = \mathop {\lim }\limits_{n \to \infty } \int_0^n {{{\left( {1 - \frac{m}{n}} \right)}^n}} \log mdm  \cr 
  &  - \gamma  = \int_0^\infty  {{e^{ - m}}} \log mdm \cr} $$
But
$$\Gamma '\left( n \right) = \int_0^\infty  {{e^{ - m}}} {m^{n - 1}}\log mdm \Rightarrow \Gamma '\left( 1 \right) = \int_0^\infty  {{e^{ - m}}} \log mdm$$
A: $\Gamma'(x)=\Gamma(x)\psi(x)$
where $\psi(x)$ is the digamma function, i.e. the logarithmic derivative of $\Gamma(x)$.  Then
$$
\lim_{x \to \infty}\frac{\Gamma'(x+1)}{x!}-\log(x)=
\lim_{x \to \infty}\frac{\Gamma(x+1)\psi(x+1)}{\Gamma(x+1)}-\log(x)=
\lim_{x \to \infty}\psi(x+1)-\log(x)
$$
Noting 
$$\psi(x+1) = \log x+O\left(\frac{1}{x}\right)$$
We conclude
$$\lim_{x \to \infty}\psi(x+1)-\log(x)=0$$
