# Infinite Sequences

• June 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??
• June 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 $a_n = \frac{5^n + 2}{7^n}$ or $a_n = \frac{5^{n + 2}}{7^n}\,$?

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

Originally Posted by Jessica098

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

(You know $\lim_{n \rightarrow +\infty} \left( \frac{5}{7} \right)^n = 0$, right?).
• June 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 $5^{n+2} = 5^n \times 5^2 = 5^n \times 25 = 25 (5^n)$ ? Now go back and read my previous reply again.
• June 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