# limit of a sequence

• Mar 9th 2010, 03:54 AM
Gok2
limit of a sequence
Hey people.
I have a sequence {b_n} which is defined like that :
b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

I need to prove that lim (b_n) as n->inf = inf
I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

Any clues anyone?
Thanks!
• Mar 9th 2010, 04:08 AM
Prove It
Quote:

Originally Posted by Gok2
Hey people.
I have a sequence {b_n} which is defined like that :
b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

I need to prove that lim (b_n) as n->inf = inf
I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

Any clues anyone?
Thanks!

What you have posted is not a sequence, but a series.

Are you trying to evaluate the $n^{\textrm{th}}$ term of the series, or are you trying to show that the series is divergent?
• Mar 9th 2010, 04:37 AM
Quote:

Originally Posted by Gok2
Hey people.
I have a sequence {b_n} which is defined like that :
b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

I need to prove that lim (b_n) as n->inf = inf
I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

Any clues anyone?
Thanks!

$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+ \frac{1}{11}+\frac{1}{13}+....$

Notice that

$\frac{1}{5}+\frac{1}{7}>\frac{2}{6}$

$\frac{1}{9}+\frac{1}{11}>\frac{2}{10}$

$\frac{1}{13}+\frac{1}{15}>\frac{2}{14}$

In this way, there is a relationship between partial sums,

$\sum_{n=1}^{\infty}\frac{1}{2n-1}>1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+.....$

which is diverging
• Mar 9th 2010, 07:00 AM
Gok2
Quote:

$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+ \frac{1}{11}+\frac{1}{13}+....$

Notice that

$\frac{1}{5}+\frac{1}{7}>\frac{2}{6}$

$\frac{1}{9}+\frac{1}{11}>\frac{2}{10}$

$\frac{1}{13}+\frac{1}{15}>\frac{2}{14}$

In this way, there is a relationship between partial sums,
$\sum_{n=1}^{\infty}\frac{1}{2n-1}>1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+.....$