# Show that $(a_{n})_{n=m}^{\infty}$ converges to $c$ iff $(a_{n})_{n=m'}^{\infty}$ converges to $c$.

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?

• 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. – user735816 Apr 26 at 0:46
• Indeed, my mistake. I have edited it. Can you please check if I am reasoning correctly this time? – BrickByBrick Apr 26 at 0:55
• You answer looks perfect now :) – user735816 Apr 26 at 0:57
• Thanks for the feedback – 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.