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Infinite series question

  1. Oct 21, 2012 #1
    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!

  2. jcsd
  3. Oct 21, 2012 #2


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    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.
  4. Oct 21, 2012 #3
    Is there any way to be more rigorous, i.e. using the Cauchy limit?

  5. Oct 21, 2012 #4


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    That the limit exists implies that the Cauchy condition holds.
  6. Oct 23, 2012 #5
    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].
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