Show that $a_n:=\frac{(-1)^{n-1}}{2n-1}$ converges

Question

Show that $a_n:=\frac{(-1)^{n-1}}{2n-1}$ converges.

If $(a_n)$ converges, the sequences is a Cauchy sequence. Which means:

$\forall \epsilon >0 \,\,\,\exists N \in \mathbb{N}\,\,\,m,n>N:\left|\frac{(-1)^{n-1}}{2n-1}-\frac{(-1)^{m-1}}{2m-1}\right|<\epsilon$

$\left|\frac{(-1)^{n-1}}{2n-1}-\frac{(-1)^{m-1}}{2m-1}\right|\le\left|\frac{1}{2n-1}\right|+\left|\frac{1}{2m-1}\right|<\left|\frac{1}{2N-1}\right|+\left|\frac{1}{2N-1}\right|=\frac{2}{2N-1}<\epsilon\Longleftrightarrow N>\frac{2+\epsilon}{2\epsilon}$

So we choose $N:=\left\lceil \frac{2+\epsilon}{2\epsilon}\right\rceil$

Which means we will always find a $N$ for any $\epsilon>0$

Is this correct?

Answer 1

This solution is indeed valid.

Rather than use the Cauchy condition, you could also prove that it converges to $0$ directly: for any $\varepsilon>0$, $|a_n|$ is less than $\varepsilon$ for all $n>\frac 12 \left(1 + \frac 1 \varepsilon\right)$.