# Infinite series question

1. Oct 21, 2012

### Bipolarity

This might sound like a dumb question, but it's actually not too obvious to me. If we know that $\lim_{n→∞}S_{n} = L$, can we prove that $\lim_{n→∞}S_{n-1} = L$ ? 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

2. Oct 21, 2012

### 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.

3. Oct 21, 2012

### Bipolarity

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

BiP

4. Oct 21, 2012

### arildno

That the limit exists implies that the Cauchy condition holds.

5. Oct 23, 2012

### Hawkeye18

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

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

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