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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 $m'<m$, 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?

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    $\begingroup$ 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 $m' \geq 1$). $\endgroup$ – MisterRiemann Jun 6 at 18:04
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Simply, adding or removing finite number of elements to/from a sequence doesn't affect its convergence.

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

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