# Thread: Infinite sequences and series?

1. ## Infinite sequences and series?

Find the limit as n approaches infinity

1. (n^3-2)/(n^2)

(n^3/n^2) -2/n^2)/ (n^2/n^2)

(n-2/n^2) I am not sure how to proceed now

2. (7-2n)/(5n) I managed to reduce it to

(7n-2)/5 And I end up getting limit is -2/5 is this correct

2. $\displaystyle \frac{n^3-2}{n^2} =\frac{n^3}{n^2}-\frac{2}{n^2} =n-\frac{2}{n^2}$

3. You're on your way with question 1, and pickslides showed you how it simplifies. As you can see, there is an n term that grows without bound, while the fractional term will decrease to zero as n grows.

The second one simplifies as follows:

$\frac{7n-2}{5n} = \frac{7n}{5n} - \frac{2}{5n} = \frac{7}{5} - \frac{2}{5n}$

Notice that the right-term will get smaller and smaller as n grows. Thus, the right-term will go to 0 as n goes to infinity, leaving you with 7/5 as the answer. Do you see how these problems were simplified and how to interpret the limit? The title of this thread mentioned infinite series, but you only included terms for a sequence. Are these to be interpreted as infinite series, too?

4. I have one more question well actually two question one is there any way I can reduce (n-(2/n^2) because I end up getting 0 and according to my book there is no limit.

And my second question is this I am trying to find the limit for (3n+4)(1-n)/(n^2) I went on the wolfram algebra website and it told me the limit is -3 using l hospitals rule the only thing is that I have not taken a calculus class before and I do not know that rule. Is there any way I can solve it without using it?

5. $\displaystyle \lim_{n\to \infty }\frac{(3n+4)(1-n)}{n^2}$

$\displaystyle \lim_{n\to \infty }\frac{-n-3n^2+4}{n^2}$

$\displaystyle \lim_{n\to \infty }\frac{-n}{n^2}+\frac{-3n^2}{n^2}+\frac{4}{n^2}$

$\displaystyle \lim_{n\to \infty }\frac{-1}{n}-3+\frac{4}{n^2}$

$\displaystyle \frac{-1}{\infty^2}-3+\frac{4}{\infty^2} = 0 -3+0 = -3$