Suppose that $(a_{n})_{n=m}^{\infty}$ is a convergent sequence of real numbers. Then $(a_{n})_{n=m}^{\infty}$ is also a Cauchy sequence.

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Suppose that $(a_{n})_{n=1}^{\infty}$ is a convergent sequence of real numbers. Then $(a_{n})_{n=1}^{\infty}$ is also a Cauchy sequence.

MY ATTEMPT

Since $a_{n}$ converges, let' say, to $L$, thus for every $\epsilon/2 > 0$ there exists a natural $N\geq 1$ such that $|a_{n} - L| \leq \epsilon/2$.

Assuming that $n,m\geq N$, the application of the triangle inequality results into \begin{align*} |a_{n} - a_{m}| = |(a_{n} - L) - (a_{m} - L)| \leq |a_{n} - L| + |a_{m} - L| \leq \epsilon \end{align*}

Consequently, $a_{n}$ is a Cauchy sequences, just as desired.

Can someone double-check my solution?