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I wanted to ask you is it possible to define that the number n is rational or irrational from analysis of integral form of function of series, for e. x. we have series $$\sum_{n=1}^{\infty}\frac{1}{{n^2}}$$ and we don't know that the sum is rational or irrational, (we assume that we don't know that is $\frac{π ^2}{6}$). But we can calculate the integral

$${\int_{0}^{\infty}\frac{1}{n^2}\,dn=1}$$

Can we say something about sum, if it is rational or irrational without calculating it?

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  • $\begingroup$ I don't understand your question (and have no idea about what is going on with the weird choice of tags), but rationality of a series $\sum_{n = 1}^{\infty} f(n)$ and an integral $\int_1^{\infty} f(x) \, dx$ are generally unrelated. $\endgroup$ – T. Bongers Oct 6 '18 at 21:07
  • $\begingroup$ So, if they not related, how to check if the sum is rational or not? $\endgroup$ – Marianna Kalwat Oct 6 '18 at 21:13
  • $\begingroup$ The irrationality of the sum is a consequence of the fact that $\pi$ is not an algebraic number. $\endgroup$ – T. Bongers Oct 6 '18 at 21:15
  • $\begingroup$ Yes, but it is assumed that we don't know the result of the sum. $\endgroup$ – Marianna Kalwat Oct 6 '18 at 21:20
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According to Wikipedia (which I deem trustworthy in this case), we can write the Euler-Mascheroni constant $\gamma$ as $$ \gamma=\sum_{n=1}^{\infty}\frac{|G_n|}{n} $$ where $G_n$ is the $n$th Gregory coefficient. The terms of the series are rational, but it's still unknown whether $\gamma$ is rational or irrational.

Another series expansion is $$ \gamma=\sum_{n=1}^\infty\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) $$ We could consider the integral $$ \int_1^\infty\left(\frac{1}{x}-\log\left(1+\frac{1}{x}\right)\right)\,dx=2\log2-1 $$ (if my computation is correct). This is irrational, actually transcendental, but cannot give insight on the nature of $\gamma$.

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  • $\begingroup$ It might be sound crazy but the result of your integral is $log4 - 1$ and the sum of "Alternating Euler-Mascheroni constant" is $$ \sum_{n=1}^\infty\left((-1)^{n-1} (\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right)=log\frac{4}{π}$$ It's similar. $\endgroup$ – Marianna Kalwat Oct 6 '18 at 23:39
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Short answer: no. The integral proves the sum converges by providing a bound on the values of the increasing sequence of partial sums. The fact that the bound is rational doesn't help trying to decide whether the sum is rational.

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  • $\begingroup$ Something like Parseval's identity could help to decide? $\endgroup$ – Marianna Kalwat Oct 7 '18 at 0:02

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