I have been trying to prove without success that the digamma function evaluated at $1$ is equal to the Euler-Mascheroni constant $\gamma$.

I would like to know if there is any way of doing it by just assuming the following facts:

  • the definition of the gamma and of the digamma functions;
  • the Bohr-Mollerup Theorem;
  • $\Gamma(x) = \lim \limits_{n \to \infty}\left(\frac{n!n^x}{(n+x)\cdots(x)}\right)$, $x>0$;
  • $\gamma = \lim \limits_{n \to \infty}\left(\sum_{1}^{n}1/n - \log n\right)$.

My attempt so far consisted in noticing that the Eulero-Mascheroni limit can be recovered by taking the logarithmic derivative of the limit definition of gamma. However, I have not been able to prove uniform convergence around $1$, which would have been enough to justify swapping limits or using results on convergence of the sequence of derivatives.

As always, any comment or answer are much appreciated and let me know if I can explain myself clearer!

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Leonhard_Euler . You use the Italian spelling ;-) $\endgroup$
    – Paul Frost
    Sep 5, 2022 at 10:02
  • $\begingroup$ @PaulFrost Thank you for pointing that out! $\endgroup$ Sep 5, 2022 at 10:37
  • $\begingroup$ You could use the Weierstraß representation for the gamma function. Though I confess the only ‘elementary’ proof I have seen of this identity played fast and loose with convergence, and the Weierstraß factorisation theorem itself is advanced $\endgroup$
    – FShrike
    Sep 5, 2022 at 10:41
  • $\begingroup$ @FShrike Thank you very much for the answer! I have asked for an elementary solution because I have read somewhere that what I have assumed in my question should be enough provided one manages to interchange limits, which is exactly my problem. $\endgroup$ Sep 5, 2022 at 23:48

1 Answer 1


Get the power series for $\log\Gamma(1+x)$ using the (provided) definition of $\Gamma$ and the discrete DCT, and then take the derivative at $x=0$ (which becomes a trivial step). Writing the definition as $$\Gamma(1+x)=x\ \Gamma(x)=\lim_{n\to\infty}\frac{n!\ n^x}{(1+x)\cdots(n+x)}=\lim_{n\to\infty}n^x\prod_{k=1}^n\frac{k}{k+x},$$ we get, for $|x|<1$, \begin{align} \log\Gamma(1+x)&=\lim_{n\to\infty}\left[x\log n-\sum_{k=1}^n\log\left(1+\frac xk\right)\right] \\&=\lim_{n\to\infty}\left[x\log n+\sum_{k=1}^n\sum_{j=1}^\infty\frac1j\left(-\frac xk\right)^j\right] \\&=\lim_{n\to\infty}\left[x\left(\log n-\sum_{k=1}^n\frac1k\right)+\sum_{j=2}^\infty\frac{(-x)^j}{j}\sum_{k=1}^n\frac1{k^j}\right] \\&=-\gamma x+\sum_{j=2}^\infty\frac{(-x)^j}{j}\zeta(j), \end{align} with the last line by Tannery (i.e. discrete DCT).


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