What is the convergence status of the given sequence and series?

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The sequence defined by An = 5n/(12n+5) converges to 5/12 as n approaches infinity. For the series from n=1 to infinity of An, it diverges because the terms do not approach zero; instead, they approach 5/12. The absence of a common ratio indicates that the series cannot converge. Therefore, the series is classified as divergent. In conclusion, the sequence converges while the series diverges.
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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 don't 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
 
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ILoveBaseball said:
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 don't 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


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 +. . .
 
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
 
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.
 
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