0

Suppose $\{a_n\}$ is a sequence of real numbers that converges to $0$ but whose series diverges. Show that for amy real number $r$ there is a sequence $\sigma_n$, consisting of $1$ and $-1$'s, such that $$\sum_n \sigma_n a_n = r$$.

I am confused about how to show this. Considering the partial sums, we can construct a sequence of partial sums such that the sum of first few terms (by making all of them positive) go just above $r$, and make the 'second' frw terms all negative so that the partial sums up to that point goes just below $r$, and continue the process. But I wasn't sure how to make this mathematically rigorous.

Squirrel-Power
  • 3,638
  • This has already been asked. Check https://math.stackexchange.com/questions/148663/if-a-n-goes-to-zero-can-we-find-signs-s-n-such-that-sum-s-n-a-n-converge – Francisco Maion Dec 12 '17 at 17:53
  • @FranciscoMaion This is not a duplicate. We are given a real sequence $a_n$ here. The complex case is much more difficult – zhw. Dec 12 '17 at 18:19

1 Answers1

0

Hint: You're on the right track. We just need to choose these subsums carefully.

Let $N_1$ be the smallest integer such that $\sum_{n=1}^{N_1}|a_n| > r.$ Then let $N_2$ be the smallest integer such that

$$\sum_{n=1}^{N_1}|a_n| - \sum_{n=N_1 + 1}^{N_2}|a_n| < r.$$

Continue ... Call these sums $S_1, S_2, \dots.$ The "smallest" criterion in choosing the $N_k,$ together with the fact that $a_n\to 0,$ guarantees that $S_1-S_2 + S_3 - \cdots = r.$

zhw.
  • 105,693