Necessary condition of changing signs of a divergent series $\sum_{n=1}^{\infty}p_{n}$ to make it convergent,$p_{n}$ decreases and tends to $0$.

Question

Let $p_{n}$ decreases and tends to $0$ while $\sum_{n=1}^{\infty}p_{n}$ is divergent. We choose $\varepsilon_{n}=\pm 1$ to make $\sum_{n=1}^{\infty}\varepsilon_{n}p_{n}$ convergent. I want to prove that $$\liminf_{n\to\infty}\frac{\varepsilon_{1}+\cdots+\varepsilon_{n}}{n}\leq0\leq\limsup_{n\to\infty}\frac{\varepsilon_{1}+\cdots+\varepsilon_{n}}{n}.$$ I think this is true because the number of positive terms had better be as much as the number of negative terms to make the series convergent.But I cannot prove it.Any help will be thanked.

Answer 1

Let $S_n = \epsilon_1 + \ldots + \epsilon_n$, and apply summation by parts to get$$\sum_{j=1}^n \epsilon_jp_j = S_n p_n + \sum_{j=1}^{n-1} S_j(p_j - p_{j+1}).$$

If $\liminf S_n/n = \beta > 0$, then there exists $N$ such that $S_n/n > \beta/2$ for all $n > N$. Using the hypothesis that $p_n$ decreases and is positive since it converges to $0$ we have that $p_j - p_{j+1} > 0$ for all $j$ and

$$\begin{align}\sum_{j=1}^n \epsilon_jp_j &> \frac{\beta}{2}np_n + \sum_{j=1}^{N} S_j(p_j - p_{j+1}) + \frac{\beta}{2} \sum_{j=N+1}^{n-1}j(p_j - p_{j+1}) \\ &= \frac{\beta}{2}np_n + \sum_{j=1}^{N} S_j(p_j - p_{j+1}) + \frac{\beta}{2} \sum_{j=N+1}^{n-1}(jp_j - (j+1)p_{j+1}) + \frac{\beta}{2}\sum_{j=N+1}^{n-1}p_{j+1} \end{align}$$

The second sum on the RHS is telescoping and, thus,

$$\tag{*}\begin{align}\sum_{j=1}^n \epsilon_jp_j &> \sum_{j=1}^{N} S_j(p_j - p_{j+1}) + \frac{\beta}{2}(N+1)p_{N+1} + \frac{\beta}{2}\sum_{j=N+1}^{n-1}p_{j+1}\end{align}$$

With $N$ fixed, the first two terms on the RHS of (*) remain constant, but the last sum tends to $+\infty$ as $n \to \infty$ since $\sum p_n$ diverges. This contradicts the convergence of $\sum \epsilon_n p_n$.

Thus, $\liminf (\epsilon_1 + \ldots + \epsilon_n)/n \leqslant 0$.

By a similar argument we can show that $\limsup (\epsilon_1 + \ldots + \epsilon_n)/n \geqslant 0$.

Answer 2

Oops. Read two inequalities backwards. Below there's an example such that $(\epsilon_1+\dots+\epsilon_n)/n$ does not tend to $0$; that answers what seems to me to be an interesting question, but not the question that was actually asked.

For $k=0,1,\dots$ let $$A_k=\{n:3^{2k}\le n<3^{2k+1}\}$$and $$B_k=\{n:3^{2k+1}\le n<3^{2k+2}\}.$$Note that the cardinality of $A_k$ is $$|A_k|=3^{2k}(3-1)=2\cdot3^{2k}$$and similarly $$|B_k|=2\cdot3^{2k+1}.$$Let $$p_n=\begin{cases}\frac{3^{-2k}}{k}&(n\in A_k), \\\frac{3^{-(2k+1)}}{k},&(n\in B_k).\end{cases}$$

Then $$\sum_{n\in A_k}p_n=\frac2k,$$so $\sum p_n=\infty$. Note that also $$\sum_{n\in B_k}p_n=\frac2k.$$Define $$\epsilon_n=\begin{cases}1,&(n\in A_k),\\-1,&(n\in B_k).\end{cases}$$Then $$\sum_{n\in A_k\cup B_k}\epsilon_np_n=0.$$And if $F\subset A_N\cup B_N$ we have $$\left|\sum_{n\in F}\epsilon_np_n\right|\le\sum_{n\in A_N\cup B_N}p_n=\frac4N.$$Define $$I_N=A_0\cup B_0\cup\dots\cup A_N\cup B_N.$$Any partial sum $s_M$ of $\sum\epsilon_np_n$ has the form $$s_M=\sum_{n\in I_N}\epsilon_np_n+\sum_{n\in F}\epsilon_np_n=\sum_{n\in F}\epsilon_np_n$$for some $F\subset I_{N+1}$; thus $|s_M|\le4/N$ (and $N\to\infty$ as $M\to\infty$); hence $$\sum\epsilon_np_n=0.$$

And finally, the number of $n\in I_N$ with $\epsilon_n=-1$ is three times the number of $n\in I_N$ with $\epsilon_n=1$, hence what you said about just as many $1$s as $-1$s is false, and in particular $(\epsilon_1+\dots+\epsilon_n)/n$ does not tend to $0$.

(If you want to put that last bit more formally, verify for yourself that $$\frac{\sum_{n\in A_k\cup B_k}\epsilon_n}{|A_k\cup B_k|}=-\frac12,$$hence $$\frac{\sum_{n\in I_N}\epsilon_n}{|I_N|}=-\frac12.)$$