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Math Help - Convergent/Divergent Series help!

  1. #1
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    Convergent/Divergent Series help!

    I have a couple of Q's I'm stuck on help and help would be MUCH appreciated! I have to use either the limit comparison or the comparison test only -
    Sum (n=1 to infinity) n/(2^n)
    Sum (n=1 to infinity) 1/(2 + n^0.5)
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  2. #2
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    Quote Originally Posted by amywilliams99 View Post
    I have a couple of Q's I'm stuck on help and help would be MUCH appreciated! I have to use either the limit comparison or the comparison test only -
    Sum (n=1 to infinity) n/(2^n)
    Sum (n=1 to infinity) 1/(2 + n^0.5)
    Consider:
    \frac{n}{2^n}=\frac{n}{\sqrt{2}^n}\cdot\frac{1}{\s  qrt{2}^n}
    Now I assume that you know that \frac{n}{\sqrt{2}^n}\rightarrow 0, for n\rightarrow \infty, because \sqrt{2}>1.
    Thus, there exists an n_0\in\mathbb{N}, such that \frac{n}{\sqrt{2}^n}<1, for all n>n_0, and thus the tail of your first series can be compared with the tail of the geometric serie \sum_{n=n_0+1}\frac{1}{\sqrt{2}^n}<\infty.

    As to your second series my guess is that for sufficiently large n, we will have that \frac{1}{2+\sqrt{n}}\geq \frac{1}{n}, thus you can compare it with the divergent harmonic series.
    To check whether \frac{1}{2+\sqrt{n}}\geq \frac{1}{n} holds true for sufficiently large n, you just note that n\geq 2\sqrt{n}\geq 2+\sqrt{n}, if n\geq 4.
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