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Math Help - series

  1. #1
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    series

    does this series converge or diverge?
    n.2^n/n!
    i did the limit test and simplified it to
    2^n+1/2^n
    after this point, im stuck
    thanx for the help!
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  2. #2
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    Quote Originally Posted by twofortwo View Post
    does this series converge or diverge?
    n.2^n/n!
    i did the limit test and simplified it to
    2^n+1/2^n
    after this point, im stuck
    thanx for the help!

    What about the n-th root test? \sqrt[n]{\frac{n2^n}{n!}}=\frac{\sqrt[n]{n}2}{\sqrt[n]{n!}}\xrightarrow [n\to\infty]{}0<1\Longrightarrow the series converges.

    Or the quotient test: \frac{(n+1)2^{n+1}}{(n+1)!}\,\frac{n!}{n2^n}= \frac{2}{n}\xrightarrow [n\to\infty]{}0\Longrightarrow the series converges.

    Or the comparison test: \frac{n2^n}{n!}\geq \frac{1}{(n-1)!} and since \sum^\infty_{n=1}\frac{1}{(n-1)!} converges so does our series.


    Tonio
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by tonio View Post

    or the comparison test: \frac{n2^n}{n!}\geq \frac{1}{(n-1)!} and since \sum^\infty_{n=1}\frac{1}{(n-1)!} converges so does our series.


    Tonio
    ????

    Cb
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    ????

    Cb

    Perhaps I missed something (nothing too surprising, uh? ), but what's not clear here? In fact, \sum^\infty_{n=1}\frac{1}{(n-1)!}=e ...

    Oh, I see....hehe! Of course, I used the comparisong test in the wrong direction...*sigh*

    Well, the OP will have to disregrad that one

    Tonio

    Ps I'm gonna add a disclaimer to my posts (this is serious, when I'll find out how to do it): "The poster of the above reply has not given too seriously a thought to the OP, so any mistake resulting from the answer is your own unique and complete responsibility".

    That'll make it...
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