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    Can someone help me with 2(ii)?

    do I say 1/n(n+1) < 1/n^2 ?Name:  Screen Shot 2017-11-08 at 13.00.47.png
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    (Original post by SWISH99)
    Can someone help me with 2(ii)?

    do I say 1/n(n+1) < 1/n^2 ?
    Yes. You should know that \displaystyle \sum \frac{1}{n^2} converges thus ...
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    (Original post by SWISH99)
    do I say 1/n(n+1) < 1/n^2 ?
    Yes that works and you are given that the 1/n^2 sequence converges.
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    (Original post by Notnek)
    Yes that works and you are given that the 1/n^2 sequence converges.
    \

    thanks a lot


    Any ideas on iii?
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    (Original post by SWISH99)
    \

    thanks a lot


    Any ideas on iii?
    \displaystyle \frac{1}{n^n} \leq \frac{1}{n^2} for all n \geq 1
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    (Original post by RDKGames)
    \displaystyle \frac{1}{n^n} \leq \frac{1}{n^2} for all n \geq 1
    cool thanks

    And I'm assuming this comparison can also be used in part ( iiv) as well?
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    (Original post by SWISH99)
    cool thanks

    And I'm assuming this comparison can also be used in part ( iiv) as well?
    actually never mind it wouldn't work
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    (Original post by SWISH99)
    actually never mind it wouldn't work
    No it wouldn't, but you can use the fact at the top of your sheet about convergence of \displaystyle \sum_{n=1}^{\infty} n^a on a suitable comparison to \frac{1}{\sqrt{n^3+1}}
 
 
 
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