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If $a_n$ is convergent and sequence $b_n = a_n - a_{n-1}$, then $\lim b_n = 0$

It's true because $\lim b_n = \lim (a_n - a_{n-1}) = \lim a_n - \lim a_{n-1} = 0$. The last limits are equal due to convergence of $a_n$. Is it correct proof?

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    $\begingroup$ Yes. Note the converse doesn't hold. $\endgroup$ – Pedro Tamaroff Jul 2 '13 at 18:11
  • $\begingroup$ Just another careful note that you may split $\lim(a_n - a_{n-1})$ into the two other limits also because $\{a_n\}$ converges. $\endgroup$ – Raymond Cheng Jul 2 '13 at 18:13
  • $\begingroup$ Yes. Can you see why the proof is not valid when $a_n \to \infty$? $\endgroup$ – Emanuele Paolini Jul 2 '13 at 18:15
  • $\begingroup$ It's a worthwhile beginning exercise to prove using $\epsilon-N$ definition, that if $\lim_{n\to\infty} a_n$ exists and $b_n=a_{n-1}$ then $\lim_{n\to\infty} b_n$ exists and equals the other limit. $\endgroup$ – Thomas Andrews Jul 2 '13 at 18:31
  • $\begingroup$ Yes. Alternatively, one can proof the result using only that $(a_n)$ is Cauchy (i.e. without using the limit $\lim a_n$ itself), right from the definition. $\endgroup$ – Hagen von Eitzen Jul 2 '13 at 18:32
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Another way to see this is by contradiction. If $\lim a_n-a_{n-1}=L\ne 0$, then for large enough $N$, we have $|a_n-a_{n-1}|>|L/2|$ for all $n>N$. It follows that $a_n$ does not converge.

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By definition, if $a_n$ converges to a, then

For every $\epsilon$, there exist an N, such that for all $n>N$, $|a_n-a| \leq \epsilon$.

Now, we want to prove that $\lim_{n \to \infty} a_n-a_{n-1} = 0$.

Choose an $\epsilon$, we need to show that for a $N$ great enough, if $n>N$, then $|a_n-a_{n-1}|< \epsilon$. We choose N such that for all $n>N$, $|a_{n-1}-a| <\frac{\epsilon}{2}$.

Then,

\begin{align} |a_n-a_{n-1}| &= |a_n-a+a-a_{n-1}| \\ & \leq |a_n-a| +|a-a_{n-1}| &\text{by triangle inequality}\\ & = \epsilon/2 +\epsilon/2\\ \end{align}

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Since $\{a_n\}$ is convergent, it is Cauchy. Given $\epsilon > 0$, choose $N$ large enough so that $n,m \ge N$ implies $|a_n - a_m| < \epsilon$. For $n > N$, we have $|a_{n} - a_{n-1}| < \epsilon$.

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