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 $c$.


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?

  • $\begingroup$ 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

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.

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