Convergence within a metric space with different starting points

I'm reading through Terence Tao's Real Analysis II, and he made a seemingly off-hand comment that made me pause and think.

"If $$(x^{(n)})_{n=m}^\infty$$ converges to $$x$$, then $$(x^{(n)})_{n=m'}^\infty$$ also converges to $$x$$ for any $$m'\geq m$$." In his notation, $$(x^{(n)})_{n=m}^\infty$$ is a sequence that starts at the $$m$$th index and is indexed by $$n$$.

Clearly if this statement also held if $$m', then he would've mentioned it, and yet I can't think of a solid reason that it couldn't hold. Does it have something to do with Riemann's rearrangement theorem?

• If you're only given the seqeuence $(x^{(n)})_{n=m}^\infty$, then you don't even know what $x^{(n)}$ means for $n < m$. However, if you're given a sequence whose index starts from $1$, then you are certainly right in that it the statement holds even for $m' < m$ (but $m' \geq 1$). – MisterRiemann Jun 6 at 18:04

So, it does hold with $$m' as well (provided those elements are defined).