Exercise 2.7.7

Question

\begin{enumerate}[label=(\alph*)] 
  \item Show that if $a_{n}>0$ and $\lim \left(n a_{n}ight)=l$ with $l 
eq 0$, then the series $\sum a_{n}$ diverges.
  \item Assume $a_{n}>0$ and $\lim \left(n^{2} a_{n}ight)$ exists. Show that $\sum a_{n}$ converges.
   \end{enumerate}

Ulisse Mini · Accepted Answer

Note: This is kind of like a wierd way to do a comparison with $1/n$ and $1/n^2$.
  \begin{enumerate}[label=(\alph*)] 
  \item If $\lim(na_n) = l 
e 0$ then $na_n \in (l-\epsilon, l+\epsilon)$, setting $\epsilon = l/2$ gives $na_n \in (l/2, 3l/2)$ implying $a_n > (l/2)(1/n)$. But if $a_n > (l/2)(1/n)$ then $\sum a_n$ diverges as it is a multiple of the harmonic series. (note that $a_n > 0$ ensures $l \ge 0$.)
  \item Letting $l = \lim(n^2a_n)$ we have $n^2a_n \in (l-\epsilon,l+\epsilon)$ setting $\epsilon = l$ gives $n^2a_n \in (0, 2l)$ implying $0 \le a_n \le 2l/n^2$ and so $\sum a_n$ converges by a comparsion test with $\sum 2l/n^2$.
   \end{enumerate}

Exercise 2.7.7

Answers

Comments

Exercise 2.7.7

Answers

Comments

Add answer