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    So I'm trying to find out if this is convergent or divergent.

    By the comparison test, I found out that this is divergent, but then with the divergence test, I find it's convergent. :confused:
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    Does \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}} converge or diverge?
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    (Original post by Noble.)
    Does \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}} converge or diverge?
    It's a p series with p<=1 so it diverges, and thus so does the original sum.
    That's what I did in the comparison test.


    But then when I took the limit to infinity of the original sum, I got 0.
    So wouldn't that mean it converges by the divergence test?
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    (Original post by eterna)
    It's a p series with p<=1 so it diverges, and thus so does the original sum.
    That's what I did in the comparison test.


    But then when I took the limit to infinity of the original sum, I got 0.
    So wouldn't that mean it converges by the divergence test?
    \displaystyle\lim_{n \rightarrow \infty} f(n) = 0 does not imply \displaystyle\sum_{n=0}^\infty f(n) is finite, for example, as you've just said yourself \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}} is not finite, yet \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{\sqrt{n}} = 0.
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    (Original post by Noble.)
    \displaystyle\lim_{n \rightarrow \infty} f(n) = 0 does not imply \displaystyle\sum_{n=0}^\infty f(n) is finite, for example, as you've just said yourself \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}} is not finite, yet \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{\sqrt{n}} = 0.
    oh ok thanks, you're right; \displaystyle\lim_{n \rightarrow \infty} f(n) = 0 doesn't necessarily imply convergence. Should have realized that while doing the comparison test.
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    (Original post by eterna)
    oh ok thanks, you're right; \displaystyle\lim_{n \rightarrow \infty} f(n) = 0 doesn't necessarily imply convergence. Should have realized that while doing the comparison test.
     \displaystyle\lim_{n \rightarrow \infty} f(n) = 0 does not imply \displaystyle\sum_{n=0}^\infty f(n) exists

    However,

    \displaystyle\sum_{n=0}^\infty f(n) exists implies  \displaystyle\lim_{n \rightarrow \infty} f(n) = 0

    That might be where you got confused.
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    (Original post by Noble.)
     \displaystyle\lim_{n \rightarrow \infty} f(n) = 0 does not imply \displaystyle\sum_{n=0}^\infty f(n) exists

    However,

    \displaystyle\sum_{n=0}^\infty f(n) exists implies  \displaystyle\lim_{n \rightarrow \infty} f(n) = 0

    That might be where you got confused.
    Thanks!
    I have another question\sum_{n=0}^\infty\frac{1}{n+ n^1/2}
    How would you go about showing that is convergent?
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    (Original post by eterna)
    Thanks!
    I have another question\sum_{n=0}^\infty\frac{1}{n+ n^1/2}
    How would you go about showing that is convergent?
    I assume you mean from n=1 and not from n=0, however this isn't convergent, via the comparison test. If you need a hint:

    n + \sqrt{n} \leq n + n = 2n the inequality is strict for n &gt; 1
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    (Original post by Noble.)
    I assume you mean from n=1 and not from n=0, however this isn't convergent, via the comparison test. If you need a hint:

    n + \sqrt{n} \leq n + n = 2n the inequality is strict for n &gt; 1
    ah yeah, I meant from n=1. thanks!
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    (Original post by eterna)
    Thanks!
    I have another question\sum_{n=0}^\infty\frac{1}{n+ n^1/2}
    How would you go about showing that is convergent?
    Are you sure that converges???
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    (Original post by davros)
    Are you sure that converges???
    It doesn't.

    \displaystyle\sum_{n=1}^\infty \frac{1}{n+ \sqrt{n}} \geq \sum_{n=1}^\infty\dfrac{1}{2n} = \dfrac{1}{2} \sum_{n=1}^\infty\dfrac{1}{n}

    So by comparison, it diverges.
 
 
 
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