Exercise 3.6

Question

Exercise 6:
    Investigate the behavior (convergence of divergence) of $\sum a_n$ if
    \begin{itemize}
        \item[(a)] $a_n=\sqrt{n+1}-\sqrt{n}$,
        \item[(b)] $a_n=\displaystyle\frac{\sqrt{n+1}-\sqrt{n}}{n}$,
        \item[(c)] $a_n=\bigl(\sqrt[n]{n}-1\bigr)^n$,
        \item[(d)] $a_n=\displaystyle\frac{1}{1+z^n}$, for complex values of $z$.
    \end{itemize}

ghostofgarborg · Accepted Answer

(a) Since the series telescopes, $S_n = \sum_{i=0}^n (\sqrt{i+1} - \sqrt i) = \sqrt{n+1}$.
    This shows that it diverges to $\infty$.

(b) 
    $$  a_n =  \frac{1}{n(\sqrt {n+1} + \sqrt n)} <  \frac{1}{n \sqrt n} $$
    Converges by the comparison test (thm. 3.25).

(c) 
    $$ \limsup_{n \to \infty} \sqrt[n]{|a_n|} = \limsup_{n \to \infty} (\sqrt[n]{n} - 1) = 1 - 1 = 0 $$
    Converges by the root test.

(d) When $|z| \leq 1$
    $$ \left| \frac{1}{1 + z^n} \right| \geq \frac{1}{1+|z|^n} \geq \frac 1 2 $$
    violating the necessary condition that $a_n \to 0$. In this case, $\sum a_n$ 
    is divergent. When $|z| > 1$, 
    $$ \left| \frac{1}{1 + z^n} \right| \leq \frac{1}{|z|^n - 1} \leq \frac{1}{|z|^n - |z|^{n-1}} = \left( \frac{|z|}{|z|-1} \right) \frac{1}{|z|^n} $$
    and $\sum a_n$ converges by comparison with the geometric series $\frac {|z|}{|z|-1} \sum \frac{1}{|z|^n}$.

Exercise 3.6

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Exercise 3.6

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