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Math Help - infinite sequences and series

  1. #1
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    infinite sequences and series

    Can someone help me with these problems Thanks!

    1. Determine whether the sequence converges or diverges, if converges, find limit. an = n/1+ squareroot n

    I divided top and bottom by 1/n and got 1/0 + 1/0 = 0 = converges

    2. find the sum of series e^n/3^n-1 I got no idea what to do, please help me, thanks again!
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  2. #2
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    Quote Originally Posted by jsu03 View Post
    Can someone help me with these problems Thanks!

    1. Determine whether the sequence converges or diverges, if converges, find limit. an = n/1+ squareroot n

    I divided top and bottom by 1/n and got 1/0 + 1/0 = 0 = converges

    2. find the sum of series e^n/3^n-1 I got no idea what to do, please help me, thanks again!

    For the first one

    \frac{n}{1+\sqrt{n}}=\frac{1}{\frac{1}{n}+\frac{1}  {\sqrt{n}}}

    Since the denominator goes to zero asn to infity the sequence divereges to infinity.

    \sum_{n=0}^{\infty}\frac{e^n}{3^{n-1}} =3\sum_{n=0}^{\infty}\frac{e^n}{3^{n}}

    This is a geometric series. I hope this helps.
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  3. #3
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    For Q2, write out the first few terms of the sequence (ie let n=1, then n=2, etc) You'll see a pattern. What sort of sequence is it? Hint: what is a? r?. Then use the relevant formula to find the sum.
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  4. #4
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    I got a= e/3, and r= 1/3 then, a/1-r = e/2 > 1 so the series diverges, is this even correct cause it look wrong to me.
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  5. #5
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    a=e not e/3 using the formula you gave.
    The sequence is (e^1)/1, (e^2)/3, (e^3)/9, (e^4)/27, ...
    ie a = e, r = e/3
    Then use a/(1-r) because r<1.
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