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Looking to get some help with constructing sequence $a_n$ of rational numbers such that $\sum |a_n|$ converges to a rational number but $\sum a_n$ does not.

My thinking is that since one sequence has absolute value then I need to make a sequence that is alternating back and forth.

would the sequence $a_n = (-1)^n$ work? as $\sum |a_n| = 1 $ but $\sum a_n$ would be bouncing back and forth between $1$ and $-1$?

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    Try for example $a_n = \frac{(-1)^n}{n(n+1)}$. The sum of $a_n$ can be evaluated using the Taylor series for $\log(1+x)$. The sum of $|a_n|$ is telescoping. – 2017-01-27
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    @Winther when evaluating that taylor series does $a_n$ diverge? and because $|a_n|$ is telescoping it converges? – 2017-01-27
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    No both series converges. In general if $\sum |a_n|$ converges then so does $\sum a_n$. – 2017-01-27
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    @Winther I was looking for an example of a sequence that the $\sum |a_n|$ converges to a rational number but $\sum a_n$ does not. – 2017-01-27
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    I see, I misunderstood. I though you meant converging to not a rational number (i.e. an irrational number). What you are asking for is impossible. As I pointed to above if $\sum |a_n|$ converges then so does $\sum a_n$. See https://en.wikipedia.org/wiki/Absolute_convergence – 2017-01-27
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    @Winther well if it converged to an irrational number what series could I use? – 2017-01-27
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    The example I gave above: $\sum |a_n|$ converges to a rational number while $\sum a_n$ converges to an irrational number. – 2017-01-27
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    @Winther say $|a_n|$ converges to a rational number, but $a_n$ does not, and converges to an irrational number – 2017-01-27
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    okay awesome thanks – 2017-01-27

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I don't think your example works: $\sum{|(-1)^n|} = |-1| +|1|+|-1|+... = 1+1+1+...$ which diverges. In fact, I don't think $\sum{|a_n|}$ can ever converge if $\sum{a_n}$ doesn't, for the following reasoning: Assume $\sum{a_n}$ diverges to positive infinity; then taking the absolute values of each term can only make the sum larger, i.e. $\sum{|a_n|} > \sum{a_n}$ if for $\sum{a_n} > 0$, and thus $\sum{|a_n|}$ diverges. If it diverges to a negative value, simply negating the sum makes it diverge to a positive value, and taking the absolute value still diverges.