Math Help - Tell if the sequence is bounded or not

1. Tell if the sequence is bounded or not

Tell if the sequence is bounded or not (indicate a reason):

$a_n=1+\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3n+1 }$

2. Re: Tell if the sequence is bounded or not

My solution:

$a_n>\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\cdots+$ $\frac{1}{3n+3}$

$=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n+1}\right)$

$=\frac{H_{n+1}}{3}$

Since the harmonic series is unbounded, $\{a_n\}$ is unbounded.

3. Re: Tell if the sequence is bounded or not

I agree. The series $\sum_{k=0}^\infty} \frac{1}{3k + 3} = \lim_{n \to \infty} \sum_{k = 0}^n \frac{1}{3k + 3} = \lim_{n \to \infty} a_n$

certainly diverges and hence the sequence is unbounded.

4. Re: Tell if the sequence is bounded or not

As explained in...

http://www.mathhelpforum.com/math-he...-i-188482.html

... is...

$a_{n}=1+\sum_{k=1}^{n} \frac{1}{3n+1}= 1+ \frac{1}{3}\ \sum_{k=1}^{n} \frac{1}{n+\frac{1}{3}}=$

$= 1-\frac{1}{3}\ \phi(\frac{1}{3}) + \frac{1}{3}\ \phi(n+\frac{1}{3}) \sim 1-\frac{1}{3}\ \phi(\frac{1}{3}) + \frac{1}{3}\ \ln (n+\frac{1}{3})$ (1)

... where...

$\phi(x)= \frac {d}{dx} \ln x!$ (2)

Kind regards

$\chi$ $\sigma$