Infinite Sequences

• Jun 8th 2008, 10:13 PM
Jessica098
Infinite Sequences
I'm hoping someone can help me out with a couple problems. For both of them, I need to determine if the sequence is convergent or divergent, and if it is convergent, I need to find the limit:

1. a(sub n) = (5^n+2)/(7^n)

2. a(sub n) = sqrt of (n+1)/(9n+1)

Anyone have any ideas or suggestions??
• Jun 8th 2008, 10:57 PM
mr fantastic
Quote:

Originally Posted by Jessica098
I'm hoping someone can help me out with a couple problems. For both of them, I need to determine if the sequence is convergent or divergent, and if it is convergent, I need to find the limit:

1. a(sub n) = (5^n+2)/(7^n)

2. a(sub n) = sqrt of (n+1)/(9n+1)

Anyone have any ideas or suggestions??

1. Is it $\displaystyle a_n = \frac{5^n + 2}{7^n}$ or $\displaystyle a_n = \frac{5^{n + 2}}{7^n}\,$?

2. Note that $\displaystyle \frac{n+1}{9n+1} = \frac{1 + \frac{1}{n}}{9 + \frac{1}{n}}$ .......
• Jun 9th 2008, 05:05 AM
Jessica098
• Jun 9th 2008, 05:10 AM
mr fantastic
Quote:

Originally Posted by Jessica098

Then $\displaystyle a_n = 25 \, \left( \frac{5}{7} \right)^n$ ......

(You know $\displaystyle \lim_{n \rightarrow +\infty} \left( \frac{5}{7} \right)^n = 0$, right?).
• Jun 9th 2008, 05:27 AM
mr fantastic
Quote:

Originally Posted by Jessica098

Yes. I know. I read it the first time.

Do you understand that $\displaystyle 5^{n+2} = 5^n \times 5^2 = 5^n \times 25 = 25 (5^n)$ ? Now go back and read my previous reply again.
• Jun 9th 2008, 06:09 AM
topsquark
Quote:

Originally Posted by Jessica098

Perhaps it would be more efficient for you if, instead of simply repeating yourself, you told Mr. F. what exactly it is about his help you are not able to follow.

-Dan