Prove that it is a cauchy sequence

Question

Show that the sequence $\langle f_n\rangle$ where

$$f_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots + \frac{(-1)^{n-1}}{n}$$

is a cauchy sequence.

My Approach: I tried solving it by starting it with $|f_n -f_m|$ but could not show this value less than any value (in terms of $m$).

Answer 1

You have \begin{align} |f_n-f_m|&=\frac1n-\frac1{n+1}+\frac1{n+2}-\frac1{n+3}+\frac1{n+4}-\cdots \\ \ \\ &=\frac1n-\left(\frac1{n+1}-\frac1{n+2}\right)-\left(\frac1{n+3}-\frac1{n+4}\right)-\cdots\\ \ \\ &=\frac1n-\frac1{(n+1)(n+2)}-\frac1{(n+3)(n+4)}-\cdots\\ \ \\ &=\frac1n-\sum_{j=n}^{m-n-1}\frac1{(n+j)(n+j+1)} \end{align} The series satisfies $$ \sum_{j=1}^{m-n-1}\frac1{(n+j)(n+j+1)}\leq\sum_{j=n}^{m-1}\frac1{n^2}.$$ So, for $n,m$ big enough, we can make $|f_n-f_m|$ arbitrarily small.

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Note: this is an ad-hoc argument. But, in general, if you have an alternating series $\sum_n(-1)^na_n$ with $a_n$ positive and decreasing to zero, we have $$ \sum_{k=n}^\infty (-1)^ka_k\leq a_n. $$

Answer 2

\begin{align} & \left| \left( 1 - \frac 1 2 + \frac 1 3 - \cdots - \frac 1 {20} \right) - \left( 1 - \frac 1 2 + \frac 1 3 - \cdots + \frac 1 {15} \right) \right| \\[10pt] = {} & \left|\frac 1 {15} - \frac 1 {16} + \frac 1 {17} - \frac 1 {18} + \frac 1 {19} - \frac 1 {20} \right| \le \frac 1 {15} \end{align} $$ \text{and more generally } \left| f_m - f_n \right| \le \frac 1 m \quad \text{ if } \quad m < n. $$ Start with $1/15,$ then subtract less than $1/15$ from it, then add less than what you subtracted, then subtract less than what you just added, and so on, and you see why the bottom line must be less than $1/15,$ and you see why this must hold more generally. And that implies you have a Cauchy sequence.