Can we Prove \lim_{n→∞}S_{n-1} = L Given \lim_{n→∞}S_{n} = L?

[Bipolarity]

This might sound like a dumb question, but it's actually not too obvious to me. If we know that [itex] \lim_{n→∞}S_{n} = L [/itex], can we prove that [itex] \lim_{n→∞}S_{n-1} = L [/itex] ? I'm actually using this as a lemma in one of my other proofs (the proof that the nth term of a convergent sum approaches 0), but can't get around the proof of this not-so-obvious-but-still-quite-intuitive lemma.

I wrote down the Cauchy-definitions of both these limits, but have no idea how to deduce one from the other.

Thanks for all the help!

BiP

 

[chiro]

Hey Bipolarity.

This seems like a bit of symbol shuffling but let m = n-1 and then consider lim m->infinity instead of lim n->infinity. Maybe I'm missing something though with what you are asking, but the nature of infinity should make the above hold.

 

[Bipolarity]

Hey Bipolarity.

This seems like a bit of symbol shuffling but let m = n-1 and then consider lim m->infinity instead of lim n->infinity. Maybe I'm missing something though with what you are asking, but the nature of infinity should make the above hold.

Is there any way to be more rigorous, i.e. using the Cauchy limit?

BiP

 

[arildno]

That the limit exists implies that the Cauchy condition holds.

 

[Hawkeye18]

If you want a rigorous proof, using ε--N definition of limit, here it is.

If [itex]\lim_{n\to\infty} S_n =L[/itex], it means that for any [itex]\varepsilon>0[/itex] there exists [itex]N=N(\varepsilon)<\infty[/itex] such that for all [itex]n>N[/itex] the inequality [itex]|S_n-L|<\varepsilon[/itex] holds.

So, for any [itex]\varepsilon>0[/itex] the inequality [itex]n>N(\varepsilon)+1[/itex] implies that [itex]n-1>N(\varepsilon)[/itex], so [itex]|S_{n-1}-L|<\varepsilon[/itex], which means that [itex]\lim_{n\to\infty} S_{n-1} =L[/itex].