# 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.

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See this as well. – J. M. Aug 16 '12 at 23:44

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 $\lim_{n\to\infty}\log(n+1)-\log(n)=0$.

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@C.Williamson $\Gamma(n+2) = (n+1) \Gamma(n+1)$, hence $\log\Gamma(n+2) = \log\Gamma(n+1) + \log(n+1)$. – Sasha Aug 16 '12 at 23:30
Yeah, I deleted my comment after seeing my mistake. I like the answer! – C. Williamson Aug 16 '12 at 23:32
Wow, this is a really clever use of the MVT, thanks. – esproff Aug 16 '12 at 23:35

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

$$\Gamma '\left( n \right) = \int_0^\infty {{e^{ - m}}} {m^{n - 1}}\log mdm \Rightarrow \Gamma '\left( 1 \right) = \int_0^\infty {{e^{ - m}}} \log mdm$$

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$\Gamma'(x)=\Gamma(x)\psi(x)$

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

$$\lim_{x \to \infty}\psi(x+1)-\log(x)=0$$

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