Show that $(s_n)$ is a Cauchy sequence

Question

$$S_n = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ... + \frac{(-1)^{n+1}}{2n-1} $$ Show that $(S_n)$ is a Cauchy sequence and hence that it converges to limit $L$. Show that $\frac{2}{3} < L < \frac{13}{15}$

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The question has been asked here

However, the question was not answered in a way that was simple for me to understand. Additionally, in my case, I must prove that the sequence is Cauchy first. In other words, I cannot show convergence and then say use Cauchy's Criterion.

Like the original question, I am stuck at let $n = m+k$

$$|S_n - S_m| = \left|\frac{(-1)^{m+2}}{2(m+1)-1}+ ... + \frac{(-1)^{m+k+1}}{2(m+k)-1}\right|$$

Now I don't know where to go from here.

Answer 1

Taking from where you leave it:

$$\left|\frac{(-1)^{m+2}}{2(m+1)-1}+ ... + \frac{(-1)^{m+k+1}}{2(m+k)-1}\right|\le\frac{k}{2(m+1)-1}\xrightarrow[m\to\infty]{}0$$

Answer 2

You can easily prove with the Alternating series test that $(S_n)$ converge, therefore it's a Cauchy sequence.