I am self-learning Real Analysis from Understanding Analysis by Stephen Abott. I'd like to ask, if my proof to the below question on convergence of infinite series is rigorous and sufficient, and checks out.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
Give an example of each or explain why the request is impossible referencing the proper theorem(s).
(a) Two series $\sum {x_n}$ and $\sum{y_n}$ that both diverge but where $\sum x_n y_n$ converges.
(b) A convergent series $\sum x_n$ and a bounded sequence $(y_n)$ such that $\sum x_n y_n$ diverges.
(c) Two sequences $(x_n)$ and $(y_n)$ where $\sum x_n$ and $\sum (x_n + y_n)$ both converge but $\sum y_n$ diverges.
(d) A sequence $(x_n)$ satisfying $0 \le x_n \le 1/n$ where $\sum (-1)^n x_n$ diverges.
Proof.
(a) The simplest examples I could come up with are:
(i) $\sum x_n = \sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum y_n = \sum_{n=1}^{\infty}\frac{1}{n}$ are both divergent sequences, but $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent.
(ii) $\sum x_n = \sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum y_n = \sum_{n=1}^{\infty}\frac{1}{n+1}$ are both divergent sequences, but $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent.
To see that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent, we simply note that $\sum_{n=1}^{\infty}\frac{1}{n^p}$ is convergent for $p > 1$, so $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent. And $\frac{1}{n(n+1)} < \frac{1}{n^2}$, so by the Comparison test, $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent.
(b) I think that this request is impossible. The series $\sum x_n$ is convergent. By the Cauchy Criterion, given any $\epsilon > 0$, there exists $N \in\mathbf{N}$, such that
\begin{align*}
\absval{x_{m+1} + x_{m+2} + \ldots + x_{n}} < \frac{\epsilon}{M}
\end{align*}
for $n > m \ge N$.
Also, the sequence $(y_n)$ is bounded, so $\absval{y_n} \le M$ for all $n\in\mathbf{N}$.
Consider the expression $\absval{x_{m+1}y_{m+1} + \ldots + x_{n}y_{n}}$. We can write,
\begin{align*}
\absval{x_{m+1}y_{m+1} + \ldots + x_{n}y_{n}} &\le \absval{x_{m+1}\absval{y_{m+1}} + \ldots + x_{n}\absval{y_{n}}}\\
&\le \absval{x_{m+1}M + \ldots + x_{n}M}\\
&=M \absval{x_{m+1} + \ldots + x_{n}}\\
&<M \cdot \frac{\epsilon}{M} = \epsilon
\end{align*}
So, by the Cauchy criterion, $\sum x_n y_n$ is a convergent series.
(c) This request is impossible. By the Algebraic Limit Theorem, if $\sum (x_n + y_n)$ converges and $\sum y_n$ converges, then $\sum (x_n + y_n) – \sum (y_n) = \sum x_n$ is also convergent.
(d) I think that this request is impossible as well. We know that $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converges. I have a hunch, that the the alternating series $\sum_{n=1}^{\infty}(-1)^n x_n$ is always bound by $\sum_{n=1}^{\infty}(-1)^n /n$.
By the Algebraic Limit theorem, $\lim_{n \to \infty} 0 \le \lim_{n \to \infty} \le \lim_{n \to \infty} \frac{1}{n}$, so $\lim_{n \to \infty}x_n = 0$. I would like to show that $(x_n)$ is a decreasing sequence.
Best Answer
(a) What you did is fine.
(b) $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n$ converges, but $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n\times(-1)^n$ diverges.
(c) What you did is fine.
(d) Take$$x_n=\begin{cases}\frac1n&\text{ if $n$ is odd}\\2^{-n}&\text{ otherwise.}\end{cases}$$