Series of non-zero rational terms converging to a rational number? On a recent test, I was asked if the following statement was true or false:
Every convergent series of non-zero rational terms converges to an irrational number
I marked it as false because I don't see why this would be true, but I couldn't think of a counterexample. Is this statement True or False? Can you provide a proof or counter example?
 A: The statement is false.
Counter-example:
$$\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$$
$$ = \sum_{n = 1}^{\infty} \frac{1}{n} - \frac{1}{n+1}$$
$$= 1.$$
Or a simpler counter example: any geometric series with $|r| < 1$.
A: Just for fun, here’s another way of constructing counterexamples.
Let $\langle r_k:k\in\Bbb N\rangle$ be any sequence of non-zero rationals converging to $0$. For $k\in\Bbb N$ let $a_{2k}=r_k$ and $a_{2k+1}=-r_k$; clearly $a_k$ is rational for all $k\in\Bbb N$. Moreover, for $n\in\Bbb N$ we have
$$\sum_{k=0}^{2n}a_k=r_n\qquad\text{and}\qquad\sum_{k=0}^{2n+1}a_k=0\;,$$
so 
$$\lim_{n\to\infty}\sum_{k=0}^na_k=0\;,$$
and $\displaystyle\sum_{k\ge 0}a_k=0$ is rational.
The idea admits any number of variations. For instance, if $\langle p_k:k\in\Bbb N\rangle$ and $\langle q_k:k\in\Bbb N\rangle$ are bounded sequences of positive rationals, you can let 
$$\begin{align*}
a_{3k}&=p_kr_k\;,\\
a_{3k+1}&=q_kr_k\;,\text{ and}\\
a_{3k+2}&=-(p_k+q_k)r_k
\end{align*}$$ 
for each $k\in\Bbb N$, and you’ll still have $\sum_{k\ge 0}a_k=0$.
