Exercise 2.7.4

Question

Give an example of each or explain why the request is impossible referencing the proper theorem(s).
  \begin{enumerate}[label=(\alph*)] 
  \item Two series $\sum x_{n}$ and $\sum y_{n}$ that both diverge but where $\sum x_{n} y_{n}$ converges.
  \item A convergent series $\sum x_{n}$ and a bounded sequence $\left(y_{n}\right)$ such that $\sum x_{n} y_{n}$ diverges.
  \item Two sequences $\left(x_{n}\right)$ and $\left(y_{n}\right)$ where $\sum x_{n}$ and $\sum\left(x_{n}+y_{n}\right)$ both converge but $\sum y_{n}$ diverges.
  \item A sequence $\left(x_{n}\right)$ satisfying $0 \leq x_{n} \leq 1 / n$ where $\sum(-1)^{n} x_{n}$ diverges.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $x_n = 1/n$ and $y_n = 1/n$ have their respective series diverge, but $\sum x_ny_n = \sum 1/n^2$ converges since it is a p-series with $p > 1$.
  \item Let $x_n = (-1)^n/n$ and $y_n = (-1)^n$. $\sum x_n$ converges but $\sum x_ny_n = \sum 1/n$ diverges.
  \item Impossible as the algebraic limit theorem for series implies $\sum(x_n+y_n) - \sum x_n = \sum y_n$ converges.
  \item The sequence
  $$x_n = \begin{cases}
      \frac{1}{n} \text{ if } n \text{ even} \
      0 \text{ otherwise}
  \end{cases}$$
   \end{enumerate}
   diverges for the same reason the harmonic series does.

Exercise 2.7.4

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Exercise 2.7.4

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