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I have some ideas about this question but I'm not sure it's right.
Heres the question:

Suppose $(x_n)_{ n\ge1}$ is a sequence of real numbers converging to $x$. Define a sequence $(y_n)_{ n\ge1}$ by $$y_n = x_{n+1} - x_n $$ Prove that the sequence $(y_n)_{ n\ge1}$ is convergent and find its limit.


This is what I know:

$x_n$ converges to $x$ so therefore $|x_n - x| <\epsilon$ such that $\epsilon > 0$.

My idea was that I could use induction to prove that $x_{n+1}$ is also convergent, but I'm not quite sure if that's right.

Any help or hints would be greatly appreciated!

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Hint: If $x_n \to x$ as $n \to \infty$, then by the triangle inequality $$\left|y_n\right| = \left|x_{n+1}-x_n\right|\le\left|x_{n+1}-x\right|+\left|x-x_n\right|$$


A few points on your approach:

  1. Be careful with your definition of convergence. $x_n \to x$ means that for any $\epsilon >0$, we can find some integer $N$ such that for every $n \ge N$, $|x_n-x|<\epsilon$. The key thing is that you don't get to pick the value of $\epsilon$: for a sequence to converge, we must be able to choose an $N$ for any $\epsilon$.
  2. If $x_n$ converges, then $x_{n+1}$ must also converge, and to the same limit (can you prove this directly from the definition of convergence?). This gives another way to prove that $y_n \to 0$, since as $n \to \infty$, $$y_n = x_{n+1}-x_n \to x-x = 0$$since if $(x_n)$ and $(y_n)$ converge, then $\displaystyle\lim_{n\to \infty}(x_n + y_n)=\lim_{n\to \infty}(x_n )+\lim_{n\to \infty}( y_n)$
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  • $\begingroup$ how would you prove that xn+1 converges if xn converges by the definition? $\endgroup$ – Zoë Soriano Dec 1 '14 at 18:02

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