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I was wondering if anyone could help me with this sequences question.

I am aiming to prove that the title sequence is a Cauchy sequence in $\mathbb Q$ but am not sure how to progress past the point where $\lvert a_m -a_n\rvert < \epsilon$

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    You may want to tell us what you know, and what results you can use. Because, for instance, this sequence does *converge* in $\mathbb{Q}$. – 2017-02-28
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    Indeed the lack of context is alarming. – 2017-02-28

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Write \begin{align} a_m-a_n &=\frac{m+1}{3m+1}-\frac{n+1}{3n+1} \\[6px] &=\frac{(3mn+3n+m+1)-(3mn+3m+n+1)}{(3m+1)(3n+1)} \\[6px] &=\frac{2(n-m)}{(3m+1)(3n+1)} \end{align} so, by $|n-m|\le n+m$, $$ |a_m-a_n|\le \frac{2n}{(3m+1)(3n+1)}+\frac{2m}{(3m+1)(3n+1)} $$ Now note that $$ \frac{2n}{3n+1}=\frac{2}{3}\frac{3n}{3n+1}<\frac{2}{3} $$ and therefore $$ |a_n-a_m|<\frac{2}{3}\left(\frac{1}{3m+1}+\frac{1}{3n+1}\right) $$ Can you finish?

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Hint.

You are going to use the fact that this sequence converges in $\mathbb Q$.

So write down the definition of a convergent sequence.

Then write down the definition of a Cauchy sequence.

Can you deduce the second one from the first one?

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    A sequence converges to a point alpha if the absolute value of the difference between alpha and am in the sequence can be made infinitely smaller as m increases, whereas a cauchy sequence is a sequence is one where the absolute value of the difference between two points in a sequence can be made infinitely small. – 2017-02-28
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    @43zombiegit Yes, and a convergent sequence is always a Cauchy sequence. – 2017-02-28
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    So if I show that the limit of the sequence results in a value, have I shown that is a Cauchy sequence? – 2017-02-28
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    Yes. Every convergent sequence is Cauchy (the converse is not true in $\mathbb{Q}$, it depends on the space). If you have seen that in class, you can use it without further justification; if not, you have to prove it. – 2017-02-28
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    @ClementC. I know that in $\mathbb{R}$ a sequence is Cauchy $\iff$ it converges. So in $\mathbb{Q}$, when a sequence is Cauchy and normally converges to an irrational number, we say that it doesn't converge? – 2017-02-28
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    @Ovi Yes absolutely! And all Cauchy converges in $E$ $\iff$ $E$ is a complete metric space. – 2017-02-28