If $\sum\frac1{a_n}$ is convergent, then irrational?

$$\{a_n\}$$ is a strictly increasing sequence of positive integers such that $$\lim_{n\to\infty}\frac{a_{n+1}}{ a_n}=1$$ If $$\sum\limits_{n=1}^\infty\frac1{a_n}$$ is convergent, can one conclude that $$\sum\limits_{n=1}^\infty\frac1{a_n}$$ is an irrational number? a transcendental number?

A special case is $$\zeta(n)(n\geq2)$$. so, the question, if true, may be difficult.

Does someone suggest a counter-example? Thanks a lot!

Take $a_n = n(n+1)$. Then

$$\sum_{n=1}^\infty \frac{1}{a_n} = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1.$$

• oh,my God! Thank you very much! – Clin May 14 '14 at 21:11

Define $a_n=n(n+1)$. Then $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\lim_{n\to\infty} \frac{n^2+3n+2}{n^2+n}=1$$

And $$\sum_{n=1}^\infty\frac1{a_n}=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac 1k-\frac1{k+1}\right)=\lim_{n\to\infty}\left(1-\frac 1{n+1}\right)=1\in\Bbb Q$$

Any example such as $\frac{1}{a_n}=\frac{1}{(n+1) \cdots (n+k)}$ for $k\geq 2$ fixed yields a telescopic series whose sum is a rational number.

Even if $a_n\geq \lambda^n$ for $\lambda>1$ one may obtain a rational number as a limit (a geometric series of the type $a_n=m^n$ will do).

If $g\geq2$ is an integer, then $\sum\limits_{n=0}^{\infty} \frac{1}{g^{n^{2}}}$ and $\sum\limits_{n=0}^{\infty} \frac{1}{g^{n!}}$ are irrational

In the cases pointed out in the above post, rational approximations to the limit are too god, and this is not the case when the number you are approximating is rational.

What is most difficult is to prove irrationality of examples such as those you pose. For instance, it is known that $\zeta(3)$ is rational (Apéry), and that $\zeta(2n)$ are powers of $\pi$ times a rational number, but in general, only partial results are known (see this for a sample, Theorem 0.2, p2):

http://wain.mi.ras.ru/PS/zete_main.pdf