Prove that $x_n = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \cdots + \frac{1}{n(n+1)}$ is Cauchy.

Question

Prove that $x_n = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \cdots + \frac{1}{n(n+1)}$ is Cauchy.

Here's my attempt:

Let $m > n$ for $m,n \in \mathbb{N}$. Then

$$ \left| x_m - x_n \right| = \left| \frac{1}{m(m+1)} +\frac{1}{(m-1)m} + \cdots+ \frac{1}{(n+1)(n+2)} \right| \tag{A}$$

Define $k = m- n$ which is a fixed integer (this is the part I'm unsure about, can I say $k$ is fixed?). Then from (A)

$$\left| \frac{1}{m(m+1)} +\frac{1}{(m-1)m} + \cdots+ \frac{1}{(n+1)(n+2)} \right| < \frac{k}{(n+1)(n+2)} < \frac{k}{n^2} $$

So choosing $N(\epsilon) = \lfloor \frac{k}{\epsilon} \rfloor + 1$ , we have

$$\left| x_m - x_n \right| = \left| \frac{1}{m(m+1)} +\frac{1}{(m-1)m} + \cdots+ \frac{1}{(n+1)(n+2)} \right| < \frac{k}{ n^2} < \frac{k}{n} < \epsilon$$ For all $m >n > N(\epsilon)$.

Answer 1

hint

Observe that

$$x_n=1-1/2+1/2-1/3+...+1/n-1/(n+1)$$

$$=1-\frac {1}{n+1} $$

and

$$x_{n+p}-x_n=\frac {1}{n+1}-\frac {1}{n+p}<\frac {1}{n+1}. $$

Answer 2

You can write $x_n = (1-(1/2))+((1/2)-(1/3))+((1/3)-(1/4))+...+((1/n)-(1/(n+1))= 1- (1/(n+1))$. Now $y_n = 1/n$ is a cauchy sequence. Therefore, $\{x_n\}$ is a cauchy sequence.