Exercise 2.3.7

Question

Give an example of each of the following, or state that such a request is impossible by referencing the proper theorem(s):
  \begin{enumerate}[label=(\alph*)] 
  \item sequences $\left(x_{n}ight)$ and $\left(y_{n}ight)$, which both diverge, but whose sum $\left(x_{n}+y_{n}ight)$ converges;
  \item sequences $\left(x_{n}ight)$ and $\left(y_{n}ight)$, where $\left(x_{n}ight)$ converges, $\left(y_{n}ight)$ diverges, and $\left(x_{n}+y_{n}ight)$ converges;
  \item a convergent sequence $\left(b_{n}ight)$ with $b_{n} 
eq 0$ for all $n$ such that $\left(1 / b_{n}ight)$ diverges;
  \item an unbounded sequence $\left(a_{n}ight)$ and a convergent sequence $\left(b_{n}ight)$ with $\left(a_{n}-b_{n}ight)$ bounded;
  \item two sequences $\left(a_{n}ight)$ and $\left(b_{n}ight)$, where $\left(a_{n} b_{n}ight)$ and $\left(a_{n}ight)$ converge but $\left(b_{n}ight)$ does not.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $(x_n) = n$ and $(y_n) = -n$ diverge but $x_n + y_n = 0$ converges
  \item Impossible, the algebraic limit theorem implies $\lim (x_n + y_n) - \lim (x_n) = \lim y_n$ therefore $(y_n)$ must converge if $(x_n)$ and $(x_n + y_n)$ converge.
  \item $b_n = 1/n$ has $b_n 	o 0$ and $1/b_n$ diverges. If $b_n 	o b 
e 0$ then $1/b_n 	o 1/b$, but since $b = 0$ ALT doesn't apply.
  \item Impossible, $|b_n|$ is convergent and therefore bounded (Theorem 2.3.2) so $|b_n| \le M_1$, and $|a_n - b_n| \le M_2$ is bounded, therefore
  $$|a_n| \leq |a_n - b_n| + |b_n| \le M_1 + M_2$$
  must be bounded.
  \item $b_n = n$ and $a_n = 0$ works. However if $(a_n) 	o a$, $a 
e 0$ and $(a_nb_n) 	o p$ then the ALT would imply $(b_n) 	o p/a$.
   \end{enumerate}

Exercise 2.3.7

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Exercise 2.3.7

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