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We know of course that rational+irrational=irrational and that ln(n) is irrational for every n greater than 1. Then why should $$\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k} - \ln(n)$$ not be irrational? Bear with me for a second, of course i know that there are counterexamples of such things, like $$x_n=\log(2)-\sum_{k=1}^n\frac{(-1)^{k+1}}k$$ But every counterexample i found looks at some series, where the underlying sequence goes to 0 and the other term is constant, or where both sequences tend to 0 like $$\frac{1}{n}+\frac{\pi}{n}$$. Or just take $\frac{\pi}{n}$, of course this is a sequence of irrationals which tends to a rational number. But these are not the exact cases for the definition of the Euler-Mascheroni constant. Both terms go to infinity. So is there a counterexample for this specific case too? Or has anyone ever looked into this specific case? Couldn't find anything on it. Maybe in the special case of $\infty-\infty$ one can indeed deduce a few things, if the limit exists.

So my exact Question is: Does there exist a rational number, which is the limit of a sum of rational numbers - a sequence of irrational numbers where both tend to infinity?

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    $\begingroup$ It would be very helpful if you spelled out your "exact question" in symbols. As it stands, "limit of a sum of rational numbers - a sequence of irrational numbers where both tend to infinity" is much too vague. $\endgroup$ – Rob Arthan Feb 19 at 20:55
  • $\begingroup$ It's hard to describe. I'd like some seqences $a_n$ and $b_n$, where a_n is rational but non-constant, and $b_n$ is a sequence of irrational numbers and both tend to infinity. But of course the answer given below delivers such sequences. So i have to think about the exact Question a little further myself. Maybe it doesn't make so much sense after all. I'd really like an example that looks like the Euler-Mascheroni constant, but is clearly rational. I'll think about it a little more and will then try to give you a question that makes mathematical sense. $\endgroup$ – Luke Feb 19 at 21:21
  • $\begingroup$ In fact , I found a claimed proof that $\gamma$ is irrational with exactly the argument you described. Of course, this proof is not valid. Unfortunately, I do not know however a concrete example where the fallacy becomes apparent. $\endgroup$ – Peter Feb 20 at 8:55
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The answer to your question is yes. For example, let $\rho$ be rational, and let $a_n$ be any sequence of irrational numbers which goes to zero. If $b_n = n + \rho$, and $c_n = n + a_n$, then $b_n,c_n\to\infty$ while $b_n-c_n \to \rho$.

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  • $\begingroup$ Indeed it is an answer to my question, but it doesn't really satisfy me. Because of course $b_n-c_n = \rho-a_n$, which is again the case where one term is constant and one sequence tends to 0. $\endgroup$ – Luke Feb 19 at 20:41

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