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Math Help - Series/Sequence

  1. #1
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    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.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ch2kb0x View Post
    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}
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  3. #3
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    Ohh Ok thanks.

    What exactly is the nth term test? And for the limit, how did you get the answer of 1/4?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ch2kb0x View Post
    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}
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