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Math Help - convergence test

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    convergence test

    I have question which to show that sum(1/n)log[1+(1/n)] is convergent. I don't think i can use integral or ratio or raabe's test. I am guessing I have to use limit comparison test ??? Help me !
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    Quote Originally Posted by mathsohard View Post
    I have question which to show that sum(1/n)log[1+(1/n)] is convergent. I don't think i can use integral or ratio or raabe's test. I am guessing I have to use limit comparison test ??? Help me !
    Hint: compare to

    \frac{1}{n^2}

    and use l'Hospitals rule
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    I haven't learned l'Hospitals rule yet lol is that the only way ???
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    Quote Originally Posted by mathsohard View Post
    I haven't learned l'Hospitals rule yet lol is that the only way ???
    Well if you do the limit comparison you will need to calculate the limit

     \frac{\frac{\log\left( 1+\frac{1}{n}\right)}{n}}{\frac{1}{n^2}}=n \log \left( 1+\frac{1}{n} \right)

    This is an indeterminate form of infinity times zero. What methods have you learned to resolve these types of limits? The only other method I can think of is to use a Taylor series expansion, but that is usually covered after the topics you are learning right now.
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    Well can you show me how to do with l'Hospitals rule then???
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    Quote Originally Posted by mathsohard View Post
    Well can you show me how to do with l'Hospitals rule then???
    Here is a link to the theorem:

    L'H˘pital's rule - Wikipedia, the free encyclopedia



     \frac{\frac{\log\left( 1+\frac{1}{n}\right)}{n}}{\frac{1}{n^2}}=\frac{ \log \left( 1+\frac{1}{n} \right)}{\frac{1}{n}}

    This is of the form zero over zero so we take the derivative of the numerator and denominator to get

    \lim_{n \to \infty}\frac{\frac{1}{ 1+\frac{1}{n}}\cdot \frac{-1}{n^2}}{\frac{-1}{n^2}}=1
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