proving that $S_n$ is Cauchy.

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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To show that $(S_n)$ is Cauchy sequence,

I tried $|S_n - S_m| < \epsilon$

Let $n = m+k$

$|S_n - S_m|$

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

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

Thanks.

Answer 1

Hint : To show it is Cauchy, combine the adjacent terms in $ \left|{ S }_{ n } - { S }_{ m } \right|$ and the rest is obvious. Or by Leibniz's test.

To show that the limit falls into a interval, you need to show that the odd terms $ { S }_{ 2k+1 } $ are decreasing and the even terms $ { S }_{ 2k } $ are increasing. You can show this by simple subtraction. Then it follows naturally that the even terms are always smaller than the odd terms because they converge to the same limit. Sum up the first few terms until two consecutive terms fall into the interval and you'll get it done.