Convergence of sequence

**jackie** · September 7th 2009, 12:27 AM

I'm grateful to any help.
I'm supposed to show that $x_{n}-x_{n-1} \rightarrow 0$ as $n \rightarrow \infty$ if $\{x_n\}$ converges.
Suppose the sequence converges to x. Since $\{x_n\}$ converges, $\{x_{n-1}\}$ also converges.
This implies that for every $\epsilon \textgreater 0$ , there exist integers $N_1, N_2$ s.t $N_1 \textgreater n$ and $N_2 \textgreater n-1$ implying absolute value of $(x_n-x) \textless \epsilon$ and absolute value of $x_{n-1}-x \textless \epsilon$ .
Am I on the right track to solve this problem? Here, do I need to use the definition of limit to show that $lim (b_n -b_{n-1})=0$ as $n \rightarrow \infty$

**girdav** · September 7th 2009, 09:22 AM

We have $\left|x_n-x_{n-1}\right| = \left|x_n-l+l-x_{n-1}\right| \leq \left|x_n-l\right|+\left|x_{n-1}-l\right| \leq \frac{\epsilon}2+\frac{\epsilon}2$ for $n \geq \max\left(N_1,N_2\right)$ .

Thread: Convergence of sequence

LinkBack

Thread Tools

Display

Convergence of sequence

Similar Math Help Forum Discussions

convergence of a sequence

Does convergence of an infinite series imply convergence of the infinite sequence?

Improper Integrals, Convergence, Sequence Convergence? Ahh!

Convergence of sequence

Sequence Convergence Using Definition of Convergence

Search Tags