Can we compute all digits of the Euler-Mascheroni constant? Let $\gamma$ be the Euler-Mascheroni constant from calculus. Is there an algorithm that computes the $n$-th digit of the decimal expansion of $\gamma$ given $n$ as input?
For all we know $\gamma$ could be a rational number, it could even be something like $k/10^n$ and have a terminating decimal expansion. Do we run into difficulties in this case when trying to compute the $n$-th or later digits?
Clarification: As another example assume we have a computable strictly increasing lower bound series $a_n$ and a computable strictly decreasing upperbound series $b_n$  both converging to an unkown real number $x$: $a_n < x < b_n$ and $b_n-a_n\rightarrow0$.
Now assume that $x$ is in fact $0.5$ but we do not know this. All we see is that no matter how (finitely) many $a_n$ and $b_n$ we compute that always $a_n < 0.5 < b_n$. How can we then ever know the first digit of the decimal expansion of $x$? Ist could be $4$ or $5$.
My questions is: Could we run into a similar problem with $\gamma$? Is there a known computable function $f(n)$ that provably computes the $n$-th digit of the decimal expansion of $\gamma$ for every $n$?
Theorertically of course there is such an algorithm in any case. But can we write it down not knowing wether $\gamma$ is irrational?
 A: In practice , we would be screwed in the case the decimal expansion is terminating. Nevertheless, $\gamma$ is computable also in this case since if it has a terminating decimal expansion, it is rational and then there is an algorithm to determine all digits. That would not help us since we would not know whether this is actually the case.
The difficulty you describe is known as the equality problem. It is in general not possible to decide the equality of two expressions if they can be "complicated enough". Not sure whether the well known limit for $\gamma$ falls into this category.
That the decimal expansion of $\gamma$ terminates is even less likely than that $\gamma$ is rational, but as far as I know, we cannot rule it out.
A: Yes, if you look at the series expansion. https://en.wikipedia.org/wiki/Euler%27s_constant#Series_expansions
One can approximate the constant by something which is easy to compute $- \log(n) + \sum_{k=1}^n \frac{1}{k}$. There is also a guarantee about how fast this converges to $\gamma$ as $n$ tends to infinity. Using that guarantee you can be sure about the first digits. It is proven that the denominator would have to huge if $\gamma$ is rational.
