Tell if the sequence is bounded or not (indicate a reason):
$\displaystyle a_n=1+\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3n+1 }$
My solution:
$\displaystyle a_n>\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\cdots+$$\displaystyle \frac{1}{3n+3}$
$\displaystyle =\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n+1}\right)$
$\displaystyle =\frac{H_{n+1}}{3}$
Since the harmonic series is unbounded, $\displaystyle \{a_n\}$ is unbounded.
As explained in...
http://www.mathhelpforum.com/math-he...-i-188482.html
... is...
$\displaystyle a_{n}=1+\sum_{k=1}^{n} \frac{1}{3n+1}= 1+ \frac{1}{3}\ \sum_{k=1}^{n} \frac{1}{n+\frac{1}{3}}=$
$\displaystyle = 1-\frac{1}{3}\ \phi(\frac{1}{3}) + \frac{1}{3}\ \phi(n+\frac{1}{3}) \sim 1-\frac{1}{3}\ \phi(\frac{1}{3}) + \frac{1}{3}\ \ln (n+\frac{1}{3}) $ (1)
... where...
$\displaystyle \phi(x)= \frac {d}{dx} \ln x!$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$