# Series/Sequence

• April 22nd 2008, 06:08 PM
ch2kb0x
Series/Sequence
An = 3n / (12n+15)

Series n=1 infinity (An) (Sorry,i dont know the symbols)
Does this converge or diverge? if it diverges does it go to infinity or minus infinity? If it converges what would be the limit?

Sequence {An}
Same question.

For the series, i said that it goes to infinity, but im guessing its wrong.

FOr the sequence, converge to 0? or diverges. not sure.
• April 22nd 2008, 06:10 PM
Mathstud28
Quote:

Originally Posted by ch2kb0x
An = 3n / (12n+15)

Series n=1 infinity (An) (Sorry,i dont know the symbols)
Does this converge or diverge? if it diverges does it go to infinity or minus infinity? If it converges what would be the limit?

Sequence {An}
Same question.

For the series, i said that it goes to infinity, but im guessing its wrong.

FOr the sequence, converge to 0? or diverges. not sure.

$\sum_{n=0}^{\infty}\frac{3n}{12n+15}$...this diverges by the n-th term test $\lim_{n \to {\infty}}\frac{3n}{12n+15}=\frac{1}{4}\ne{0}$ and as I showed the lmit for the sequence is $\frac{1}{4}$
• April 22nd 2008, 06:19 PM
ch2kb0x
Ohh Ok thanks.

What exactly is the nth term test? And for the limit, how did you get the answer of 1/4?
• April 22nd 2008, 06:25 PM
Mathstud28
Quote:

Originally Posted by ch2kb0x
Ohh Ok thanks.

What exactly is the nth term test? And for the limit, how did you get the answer of 1/4?

These are the most basics of series...you should consult your text book or teacher for furhter help....n-th term test states if $\lim_{n \to{\infty}}a_n\ne{0}$ the series is divergent...and for the limit there is many ways to do it easiest is if you multiply by $\frac{\frac{1}{n}}{\frac{1}{n}}$ you get $\lim_{n \to {\infty}}\frac{3}{12+\frac{15}{n}}=\frac{1}{4}$