Exercise 2.7.2

Question

Decide whether each of the following series converges or diverges:
  \begin{enumerate}[label=(\alph*)] 
  \item $\sum_{n=1}^{\infty} \frac{1}{2^{n}+n}$
  \item $\sum_{n=1}^{\infty} \frac{\sin (n)}{n^{2}}$
  \item $1-\frac{3}{4}+\frac{4}{6}-\frac{5}{8}+\frac{6}{10}-\frac{7}{12}+\cdots$
  \item $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots$
  \item $1-\frac{1}{2^{2}}+\frac{1}{3}-\frac{1}{4^{2}}+\frac{1}{5}-\frac{1}{6^{2}}+\frac{1}{7}-\frac{1}{8^{2}}+\cdots$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Converges by a comparison test with $\sum_{n=1}^\infty \frac{1}{2^n}$.
  \item Converges by a comparison test with $\sum_{n=1}^\infty \frac{1}{n^2}$.
  \item Diverges since $(n+1)/2n = 1/2 + 1/2n$ never gets smaller then $1/2$.
  \item Grouping terms gives
    $$
    \frac{1}{n} + \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \ge \frac{1}{n}
    $$
    Which shows the subsequence $(s_{3n})$ diverges (via comparison test with the harmonic series) hence $(s_n)$ also diverge.
  \item Intuitively this should diverge since it is a mixture of $1/n$ (divergent) and $1/n^2$ (convergent). To make this rigorous examine the subsequence $(s_{2n})$
    $$
    s_{2n} = 1 - 1/2^2 + 1/3 - 1/4^2 + \dots + 1/(2n)^2
    $$
    Let $t_n = \sum_{k=1}^n \frac{1}{2k-1}$ and $v_n = \sum_{k=1}^n \frac{1}{(2k)^2}$ so $s_{2n} = t_n - v_n$.

A comparison test with the harmonic series (after some manipulation) shows that $(t_n)$ diverges, and p-series tells us $(v_n)$ converges. Therefore their difference $s_{2n} = t_n - v_n$ must diverge, which implies $(s_n)$ diverges as desired.
   \end{enumerate}

Exercise 2.7.2

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Exercise 2.7.2

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