1
$\begingroup$

Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers, let $c$ be a real number, and let $m \geq m'$ be an integer. Show that $(a_{n})_{n=m}^{\infty}$ converges to $c$ iff $(a_{n})_{n=m'}^{\infty}$ converges to $c$.

MY ATTEMPT (EDIT)

WLOG, let us assume that $m' > m$ and $(a_{n})_{n=m}^{\infty}$ converges. According to the definition of limit, for every $\varepsilon > 0$, there exists a natural number $N_{1}\geq m$ such that \begin{align*} n\geq N_{1} \Longrightarrow |a_{n} - c| < \varepsilon \end{align*}

If we take $N = \max\{N_{1},m'\}$, we conclude that, for every $\varepsilon > 0$, there exists a natural number $N\geq m'$ such that \begin{align*} n\geq N \Longrightarrow |a_{n} - c|\leq\varepsilon \end{align*} and hence $(a_{n})_{n=m'}^{\infty}$ converges to $c$.

Conversely, on the same assumption that $m' > m$, if $(a_{n})_{n=m'}^{\infty}$ converges, for every $\varepsilon > 0$, there is a natural number $N\geq m' > m$ such that \begin{align*} n\geq N \Longrightarrow |a_{n}-c|\leq\varepsilon \end{align*} from whence we conclude that $(a_{n})_{n=m}^{\infty}$ converges to $c$ as well.

Could someone please double-check my reasoning?

$\endgroup$
  • $\begingroup$ You should assume $(a_n)_{n=m}^{\infty}$ converges to $c$ to prove $(a_n)_{n=m'}^{\infty}$ converges to $c$ as well, and vice-versa. In your reasoning, you assume both are true at the same time, which is not correct. $\endgroup$ – user735816 Apr 26 at 0:46
  • $\begingroup$ Indeed, my mistake. I have edited it. Can you please check if I am reasoning correctly this time? $\endgroup$ – BrickByBrick Apr 26 at 0:55
  • $\begingroup$ You answer looks perfect now :) $\endgroup$ – user735816 Apr 26 at 0:57
  • $\begingroup$ Thanks for the feedback $\endgroup$ – BrickByBrick Apr 26 at 1:01
1
$\begingroup$

Your thinking is correct. What I feel is that the reasoning is your answer may be modified a little bit. Mine goes as follows: If $(a_n)_{n=m}^{\infty} \rightarrow c$ then for any $\epsilon >0 \exists N \geq m$ such that $ n \geq N \Rightarrow |a_n-c|<\epsilon$. Then automatically $N \geq 𝑚′$.

For the converse part you need to choose $N_1= max \{N,m\}$ and your argument follows similarly replacing $N$ by $N_1$ above.

The main idea is that convergence of a sequence depends only on it’s tail. So, if one tail converges it holds for the whole sequence as well and the limit is also the same.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.