# Thread: Calculus 2: Series/Sequences test problem

1. ## Calculus 2: Series/Sequences test problem

So, i just took my exam. I think I did really well! Except, there were two questions that I sorta blanked out on and maybe was doing something wrong.

1) The first one was find the convergence of ((3n^2+8)/(8n^2+3))^n. I used the root test and found that the limit < 1 which means that the series converges absolutely.

However, the second part to this same series was whether the sequence ((3n^2+8)/(8n^2+3))^n converges or diverges? (if converges, compute lim as n-> inf of the seq).

---> so since the series converges, i was thinking that the sequence should converge (limit of seq. exists). However, when i tried to find the limit of the sequence, the sequence goes to inf (e^inf). Thus, i said that the sequence diverges to infinity.

My Question: shouldn't the sequence converge if the series converge? Why couldn't i find a limit for the sequence then?

2) Another question was simply write sin(5x)cos(5x) with maclaurin series using the trig identity : sin(2x) = 2sin(x)cos(x).

I knew the sin(5x) and cos(5x) series with the memorized maclaurin series. However, i was having trouble rewriting sin(5x)cos(5x)... I thought it would be
2sin(5x)cos(5x) = sin(10x) ---> sin(10x)/2.

I probably rewrote it wrong with the trig identity. I completely blanked on how to use these trig identity!

2. ## Re: Calculus 2: Series/Sequences test problem

1) As $n$ gets large, you have essentially $\left(\dfrac{3}{8}\right)^n$. This limit approaches zero as $n \to \infty$.

2) That is the correct trig identity.

3. ## Re: Calculus 2: Series/Sequences test problem

awh man. I knew something was wrong for problem 1...

It just didn't make sense. I guess i computed the limit completely wrong!

4. ## Re: Calculus 2: Series/Sequences test problem

Originally Posted by lc99
awh man. I knew something was wrong for problem 1...

It just didn't make sense. I guess i computed the limit completely wrong!
You just missed a minus sign. You should have gotten $e^{-\infty}$ as the limit, which is zero.