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Math Help - [SOLVED] 2 infinite series

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] 2 infinite series

    I wasn't able to find if they converges or not :
    1) \sum_{n=1}^{+\infty} \left( \frac{n}{2n+1}\right) ^n My intuition tells me it converges.
    And 2) \sum_{n=1}^{+\infty} (\sqrt[n]{n}-1)^n. This one is going to make me crazy. I'm not sure at all if it diverges or not.
    For both I've tried the root test, integral test and for the first the ratio test but in each case I failed to get a conclusion. I need a little help (maybe the test to do, and any trick I might not see) but not the full detailed answer, please.
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by arbolis View Post
    I wasn't able to find if they converges or not :
    1) \sum_{n=1}^{+\infty} \left( \frac{n}{2n+1}\right) ^n My intuition tells me it converges.
    And 2) \sum_{n=1}^{+\infty} (\sqrt[n]{n}-1)^n. This one is going to make me crazy. I'm not sure at all if it diverges or not.
    For both I've tried the root test, integral test and for the first the ratio test but in each case I failed to get a conclusion. I need a little help (maybe the test to do, and any trick I might not see) but not the full detailed answer, please.
    1) Use the Root test <br />
\sqrt[n]{{\left( {\tfrac{n}<br />
{{2n + 1}}} \right)^n }} = \tfrac{n}<br />
{{2n + 1}} \to \tfrac{1}<br />
{2}<br />
so it converges


    2) Again, the root test. Note that: <br />
\sqrt[n]{n} - 1 = e^{\tfrac{{\ln \left( n \right)}}<br />
{n}}  - 1 = \tfrac{{\ln \left( n \right)}}<br />
{n} + \tfrac{{\ln ^2 \left( n \right)}}<br />
{{2 \cdot n^2 }} + \tfrac{{\ln ^3 \left( n \right)}}<br />
{{3! \cdot n^3 }} + ...<br />
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  3. #3
    MHF Contributor arbolis's Avatar
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    Sorry for the first one! I realize it was much easier than I thought.
    2) Again, the root test. Note that:
    Thanks! I almost got something similar, but couldn't find it. Correct me if I'm wrong, but \frac{\ln (n)}{n} \rightarrow 0, so e^{\frac{\ln (n)}{n}} \rightarrow 1, therefore e^{\frac{\ln (n)}{n}}-1 \rightarrow 0, so the series converges by the root test.
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by arbolis View Post
    Sorry for the first one! I realize it was much easier than I thought.
    Thanks! I almost got something similar, but couldn't find it. Correct me if I'm wrong, but \frac{\ln (n)}{n} \rightarrow 0, so e^{\frac{\ln (n)}{n}} \rightarrow 1, therefore e^{\frac{\ln (n)}{n}}-1 \rightarrow 0, so the series converges by the root test.
    That's right
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I wasn't able to find if they converges or not :
    1) \sum_{n=1}^{+\infty} \left( \frac{n}{2n+1}\right) ^n My intuition tells me it converges.
    And 2) \sum_{n=1}^{+\infty} (\sqrt[n]{n}-1)^n. This one is going to make me crazy. I'm not sure at all if it diverges or not.
    For both I've tried the root test, integral test and for the first the ratio test but in each case I failed to get a conclusion. I need a little help (maybe the test to do, and any trick I might not see) but not the full detailed answer, please.
    You tell me the answer, but an even easier way of doing the first one is that,

    \left(\frac{n}{2n+1}\right)^n\sim~?
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  6. #6
    MHF Contributor arbolis's Avatar
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    You tell me the answer, but an even easier way of doing the first one is that,

    \left( \frac{1}{2}\right) ^n, or I don't know. Whatever, I prefer to be rigorous, otherwise I won't do well in the final exam .
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  7. #7
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    You tell me the answer, but an even easier way of doing the first one is that,

    \left(\frac{n}{2n+1}\right)^n\sim~?
    <br />
\left( {\tfrac{n}<br />
{{2n + 1}}} \right)^n  = \left( {\tfrac{{\tfrac{1}<br />
{2}}}<br />
{{1 + \tfrac{1}<br />
{{2n}}}}} \right)^n  = \tfrac{{\left( {\tfrac{1}<br />
{2}} \right)^n }}<br />
{{\left( {1 + \tfrac{1}<br />
{{2n}}} \right)^n }}<br />

    Now <br />
{\left( {1 + \tfrac{1}<br />
{{2n}}} \right)^n } \to {\sqrt{e}}<br />
as n tends to infinity (remember that: <br />
\left( {1 + \tfrac{x}<br />
{n}} \right)^n  \to e^x <br />
)

    So <br />
\left( {\tfrac{n}<br />
{{2n + 1}}} \right)^n  \sim \tfrac{1}<br />
{{\sqrt e }} \cdot \left( {\tfrac{1}<br />
{2}} \right)^n <br />
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    \left( \frac{1}{2}\right) ^n, or I don't know. Whatever, I prefer to be rigorous, otherwise I won't do well in the final exam .
    No thats totally fine, I was just saying, you can show also pretty easily (and a little more pecularily, which is what I like) that

    \lim_{x\to\infty}\frac{\left(\frac{n}{2n+1}\right)  ^n}{\left(\frac{1}{2}\right)^n}=e^{\frac{-1}{2}}\Rightarrow\left(\frac{n}{2n+1}\right)^n\sim  {e^{\frac{-1}{2}}\left(\frac{1}{2}\right)^n}

    Which is obviously convergent.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    <br />
\left( {\tfrac{n}<br />
{{2n + 1}}} \right)^n = \left( {\tfrac{{\tfrac{1}<br />
{2}}}<br />
{{1 + \tfrac{1}<br />
{{2n}}}}} \right)^n = \tfrac{{\left( {\tfrac{1}<br />
{2}} \right)^n }}<br />
{{\left( {1 + \tfrac{1}<br />
{{2n}}} \right)^n }}<br />

    Now <br />
{\left( {1 + \tfrac{1}<br />
{{2n}}} \right)^n } \to {\sqrt{e}}<br />
as n tends to infinity (remember that: <br />
\left( {1 + \tfrac{x}<br />
{n}} \right)^n \to e^x <br />
)

    So <br />
\left( {\tfrac{n}<br />
{{2n + 1}}} \right)^n \sim \tfrac{1}<br />
{{\sqrt e }} \cdot \left( {\tfrac{1}<br />
{2}} \right)^n <br />
    Good timing
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