Question

find lim -> infinity n^3 + 2/15n^3

Answer 1

Divide through by the highest power of \(n\):
\[\lim_{n\to\infty}\frac{n^3+2}{15n^3}=\lim_{n\to\infty}\frac{1+\dfrac{2}{n^3}}{15}\]

Answer 2

i'm confused

Answer 3

\[\frac{n^3+2}{15n^3}\cdot\frac{\dfrac{1}{n^3}}{\dfrac{1}{n^3}}=\frac{\dfrac{n^3+2}{n^3}}{\dfrac{15n^3}{n^3}}=\frac{1+\dfrac{2}{n^3}}{15}\]

Answer 4

Oh okay i get it so what is the limit?

Answer 5

1/15

Answer 6

In the future, if you see a problem where the degree in the numerator and the degree of the denominator are the same, as in
\[\frac{n^2+n+1}{2n^2-3}\]
then the limit as \(n\to\infty\) will be the ratio of the coefficients of these terms. In the case of this example, the highest power of \(n\) in both numerator and denominator is 2. The coefficient in the numerator is 1, and the coefficient in the denominator is 2, so the limit for this example would be \(\dfrac{1}{2}\).

Answer 7

OKay that makes sense thank you:)

Answer 8

yw

Answer 9

Can i ask you a quick question?

Answer 10

yes

Answer 11

5^2 + 5^3 + 5^4 + 5^5... this series coverges correct?

Answer 12

@SithsAndGiggles

Answer 13

No. What you have is a geometric series with a common ratio between terms of 5. If the common ratio is not less than 1, then the series will diverge.
\[\sum_{n=0}^\infty ar^n~~\Rightarrow~~\begin{cases}\text{converges if }|r|<1\\
\text{diverges if }|r|\ge1\end{cases}\]
In this case, \(r=5\), and \(|5|=5>1\).

Answer 14

hmm so any number would be that way too?
1/8 + 1/8^2 + 1/8^3

Answer 15

this would be divergent too?

Answer 16

@SithsAndGiggles

Answer 17

No, "any" number wouldn't make such a series diverge. Check my last comment. I listed the conditions for convergence/divergence.

For the most recent series, \(\dfrac{1}{8}+\dfrac{1}{8^2}+\dfrac{1}{8^3}+\cdots\), the common ratio is \(\dfrac{1}{8}\). Is \(\left|\dfrac{1}{8}\right|\) less than or greater than 1?

Answer 18

less so convergent