# Rational or irrational sum and the integral

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?

• 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. – T. Bongers Oct 6 '18 at 21:07
• So, if they not related, how to check if the sum is rational or not? – Marianna Kalwat Oct 6 '18 at 21:13
• The irrationality of the sum is a consequence of the fact that $\pi$ is not an algebraic number. – T. Bongers Oct 6 '18 at 21:15
• Yes, but it is assumed that we don't know the result of the sum. – Marianna Kalwat Oct 6 '18 at 21:20

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 $$\gamma$$.
• 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. – Marianna Kalwat Oct 6 '18 at 23:39