# Help me!!

#### ssolaming

##### New member
I am working on sequence problems, and I am stuck with this problem.

n = 1 to ∞ (3n/(2n-1))

the expression is above and I have to use the DIRECT COMPARISON test to determine the convergence or divergence of the series.

I know that this expression diverges, and I also know that I have to find some sequence that is smaller than the one given and prove the divergence of that sequence to prove this sequence, but I don't know which one to use.
Which sequence do I have to compare that with?

#### Subhotosh Khan

##### Super Moderator
Staff member
Can you check the behavior of:

n = 1 to ∞ (2^n/(2^n -1 ))

• topsquark

#### ssolaming

##### New member
Can you check the behavior of:

n = 1 to ∞ (2^n/(2^n -1 ))
I can't find the behavior of that one either...

#### MarkFL

##### Super Moderator
Staff member
I can't find the behavior of that one either...
What if we write the $$n$$ term in the suggested series for comparison as:

$$\displaystyle a_n=\frac{2^n}{2^n-1}=\frac{1}{1-2^{-n}}$$

What is:

$$\displaystyle \lim_{n\to\infty}a_n$$ ?

• topsquark

#### Jomo

##### Elite Member
I can't find the behavior of that one either...
A necessary condition for a series to converge is for $$\displaystyle \lim_{n\to\infty}a_n$$=0. The lim for the series that Khan gave you is 1 so it diverges.
This is a bit sneaky, but you were asked to show by LCT that your series divverges. That does not mean that you can't use other methods than LCT to show that your test series diverges!

• topsquark