1
$\begingroup$

How can I show that if $a_n$ is a sequence of real numbers, then the following statements are equivalent:

  • $a_n$ has a subsequence, $a_{n_k}$, such that the summation of the subsequence converges,i.e:

$$ \sum_{k=0}^\infty a_{n_k} $$ converges

and

$$ \liminf |a_n| = 0$$

This makes some intuitive sense to me, because if a_n has a convergent subsequence, it must be bounded somehow (above or below or both), and if liminf of the absolute value sequence is zero, then it again means that the sequence must be in some way bounded, because if it didnt it would diverge to infinity.

However I have no idea where to start on this in a formal proof sense. Does anyone have any tips for how to form this connection? I don't really know anything concrete that I can conclude from the second statement (the liminf one) I feel like maybe bolzano-weierstrauss can give me a hand somehow??

$\endgroup$
1
$\begingroup$

Check first that $\liminf |a_n|=0$ if and only if some subsequence of the $a_n$ converges to $0$.

Assuming that $\liminf|a_n|=0$, use the equivalence above to show that, indeed, there is a subsequence $(a_{n_k})_{k\ge0}$ that, not only converges to 0, but does it very quickly, say $|a_{n_k}|<1/2^k$ for all $k$. Check that the sum of this subsequence converges.

For the converse, if there is a subsequence whose series converges, this subsequence converges to 0 (right?). Use the equivalence mentioned in the first paragraph to conclude.

$\endgroup$
  • $\begingroup$ How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$ $\endgroup$ – ktuggle Nov 21 '18 at 3:32
  • 1
    $\begingroup$ Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence. $\endgroup$ – Andrés E. Caicedo Nov 21 '18 at 4:02

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.