Where can I find a proof for $\lim\limits_{n \to \infty} \left (n - \Gamma \left( \frac 1n \right) \right) = \gamma$?

I found in a book the following limit: $\lim\limits_{n \to \infty} \left (n - \Gamma \left( \frac 1n \right) \right) = \gamma$.

They say that a proof for this is in "Havil, J.: GAMMA, Exploring Euler’s Constant. Princeton University Press, Princeton (2003)? (p. 109)

Unfortunately, I don't have this book, and I would like to see a proof of this limit. Can anybody help me, please?

Thank you!

• I couldn't find a proof of this in that book, but I could find that it mentions this fact in page 109 – DonAntonio Feb 20 '16 at 19:00
• Hm. You're right! They didn't say that in the book mentioned above is a proof, but I thought it is! I'm sorry! – npatrat Feb 20 '16 at 19:12

If you know $\;-\gamma=\Gamma'(1)\;$ then this is just the same:
$$n-\Gamma\left(\frac1n\right)=\frac{1-\frac{\Gamma\left(\frac1n\right)}{n}}{\frac1n}=\frac{1-\frac1n\Gamma\left(\frac1n\right)}{\frac1n}=$$
$$\frac{1-\Gamma\left(1+\frac1n\right)}{\frac1n}\xrightarrow[n\to\infty]{}-\Gamma'(1)$$
$$\lim_{n\to\infty} (n-\Gamma(1/n))=\lim_{n\to0} (1/n-\Gamma(n))=\lim_{n\to0} \frac{1-n\Gamma(n)}n=\lim_{n\to0} \frac{1-\Gamma(n+1)}n$$ Which is indeterminate so by L'Hopital $$\lim_{n\to0} (-\Gamma'(n+1))=-(-\gamma)=\gamma$$ The value for $\Gamma(1)$ can be derived from $$\frac{\Gamma'(n)}{\Gamma(n)}=H_{n-1}-\gamma$$