Q:

On convergence of variation of harmonic series with some conditions on the coefficients

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? 

A:

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 $x_k=\varepsilon_k/k.$

