Suppose that we have series $\sum_{n=1}^{\infty} \dfrac {\varepsilon_n}{n}$ where $\varepsilon_n \in \{-1,1\}$ for every $n \in \mathbb N$.

If we choose $\varepsilon_n=1$ for every $n \in \mathbb N$ then we have harmonic series which is divergent.

If we choose $\varepsilon_{2n-1}=1$ and $\varepsilon_{2n}=-1$ then we have alternating harmonic series which sums to $\ln2$.

If we denote by $p_n$ the number of positive terms in the set $\{\varepsilon_1,...,\varepsilon_n\}$, then, for the harmonic series, we have $\lim_{n\to\infty} \dfrac {p_n}{n}=1$, and for the alternating harmonic series we have $\lim_{n\to\infty} \dfrac {p_n}{n}=\dfrac {1}{2}$.

My question is:

Can we have sequence $\varepsilon_n$ such that $\lim_{n\to\infty} \dfrac {p_n}{n}>\dfrac {1}{2}$ and $\sum_{n=1}^{\infty} \dfrac {\varepsilon_n}{n}$ is convergent?

  • $\begingroup$ @Ian But in my case $\varepsilon_n$ cannot take the value $0$: $\endgroup$ – Farewell Dec 9 '15 at 1:51
  • $\begingroup$ @Ian Could it be the case that after some point the number of such minus ones becomes so large that the series start exhibit behavior towards divergence? $\endgroup$ – Farewell Dec 9 '15 at 1:57
  • $\begingroup$ I think that is the case. $\endgroup$ – Element118 Dec 9 '15 at 2:03
  • $\begingroup$ @Ian Are not both types of terms infinite in number? $\endgroup$ – Farewell Dec 9 '15 at 2:05
  • $\begingroup$ @Ian There are a lot of numbers with a $1$ in base $N$ representation, in the sense that almost all numbers have a $1$. As such, I am willing to guess that the answer is likely to be no. $\endgroup$ – Element118 Dec 9 '15 at 2:07

The answer to your question is no.

Suppose that $\sum_{n=1}^N \varepsilon_n/n\to\alpha.$ Taking Cesàro averages, we must also have ${1\over N}\sum_{k=1}^{N}\left(\sum_{n=1}^{k-1} \varepsilon_n/n\right)\to \alpha$. But this last sum can be rewritten as $${1\over N}\sum_{k=1}^{N}\left(\sum_{n=1}^{k-1} {\varepsilon_n\over n}\right)= {1\over N}\sum_{n=1}^{N}\left(\sum_{k=n+1}^{N} {\varepsilon_n\over n}\right) = {1\over N}\sum_{n=1}^{N} (N-n)\, {\varepsilon_n\over n} =\sum_{n=1}^N {\varepsilon_n\over n}-{1\over N}\sum_{n=1}^N\varepsilon_n.$$

It follows that ${1\over N}\sum_{n=1}^N\varepsilon_n\to0$ as $N\to\infty$. The proportion of plus signs and minus signs must be asymptotically equal.

This result is well-known, and follows from Kronecker's Lemma by setting $b_n=n$ and $x_k=\varepsilon_k/k.$

  • $\begingroup$ In the second sum on $n$, the upper limit should be $N-1$. Does this affect your proof? $\endgroup$ – marty cohen Dec 9 '15 at 3:29
  • $\begingroup$ @martycohen The extra term is zero, the empty sum. $\endgroup$ – user940 Dec 9 '15 at 3:30

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