# Series and sequence

1. Apr 23, 2005

### ILoveBaseball

$$An = \frac{5n}{12n+5}$$

For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise.

a.) The series n=1 to infinity(i dont know how to make the 'sum of' sign)
(An) i'm having trouble with this one, there seems to be no 'common ratio', so does this mean it's divergent? $$A_1 = 5/17, A_2 = 10/29, A_3 = 15/41$$

b.) the squence $$A_n$$. well for this one as n-> infinity, the limit should be 5/12

2. Apr 23, 2005

### OlderDan

Your b) looks good and should tell you the answer to a) What is 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 +. . .

3. Apr 23, 2005

### James R

If

$$A_n = \frac{5n}{12n+5}$$

then

$$\lim_{n \rightarrow \infty} A_n = \lim_{n \rightarrow \infty} \frac{5}{12+5/n}$$
$$=\frac{5}{12+(\lim_{n \rightarrow \infty} 5/n)} = 5/12$$

4. Apr 23, 2005

### HallsofIvy

Staff Emeritus
Hint for a) In order for a series such as $\Sigma_1^{\infinity}A_n$ to converge, it is necessary that the sequence {An} converge to 0.