# [Calc2] Question about convergence of series.

1. Oct 28, 2004

### gflores

I have this series
E(n=3) = (4n+3) / (7n-1)... going to infinity. I'm not quite sure how I would go about solving this. Do I use the comparison test? Any help would be greatly appreciated. Thank you.

2. Oct 29, 2004

### vsage

Yes use direct comparison. I suggest something like (5/6)^n or something.

3. Oct 29, 2004

### shmoe

I'm not totally clear on what your series is. is it:

$$\sum_{n=3}^{\infty}\frac{4n+3}{7n-1}$$

if so, what is $$\lim_{n\rightarrow\infty}\frac{4n+3}{7n-1}$$?

4. Oct 29, 2004

### gflores

5. Oct 29, 2004

### shmoe

Hit the "reply" button to my post above and you can see how it was done. You can use LaTex on this board, there's a quick guide kicking around somewhere.

I don't understand vsage's suggestion either. I'm guessing he interpreted the question differently?

Yes, that is correct. What does it say about the series if the limit of the terms is 4/7?

6. Oct 29, 2004

### vsage

Wow I'm sorry I wrote that way past my bedtime :\. To use the comparison test on a series $$a_n$$ to show that $$b_n$$ diverges, $$b_n \geq a_n \forall n$$. By the p test the series you gave doesn't converge. That would be the easest way to go or use what shmoe suggested that since the limit of the terms of the series don't go to 0 that the series can't converge but to use comparison you know by the p series test that $$\sum \frac{4n}{7n}$$ diverges so using $$\frac{4n}{7n} \leq \frac{4n+3}{7n-1} \forall n$$ therefore the series diverges. Again very sorry for misreading it. I saw it as $$\frac {(4n+3)^n}{(7n-1)^n}$$

Last edited by a moderator: Oct 29, 2004
7. Oct 29, 2004

### Corneo

Just use the divergence test like it was mentioned before.

The limit as you take n to inf. is 4/7 meaning as n gets bigger, your still adding more numbers and they don't get to 0. So the series diverge.