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Math Help - Test for convergence

  1. #1
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    Test for convergence

    \displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{ln(1+e^n)}

    \displaystyle \sum_{n = 1}^{\infty} (-1)^n (1-\displaystyle \frac{1}{n})^{n^2}

    known that the first one is conditionally convergent and the second one is absolutely convergent, which test shud i use to prove it
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  2. #2
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    Quote Originally Posted by wopashui View Post
    \displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{ln(1+e^n)}

    \displaystyle \sum_{n = 1}^{\infty} (-1)^n (1-\displaystyle \frac{1}{n})^{n^2}

    known that the first one is conditionally convergent and the second one is absolutely convergent, which test shud i use to prove it

    \displaystyle{\frac{1}{\ln(1+e^n)} is monotonically descending convergent to zero so \displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{\ln(1+e^n)}} is cond. convergent.

    It isn't abs. convergent since \displaystyle{\frac{1}{\ln(1+e^n)}\geq\frac{1}{\ln (2e^n)}=\frac{1}{\ln 2+n} , which is divergent.

    For the second one the root test gives \displaystyle{\left[\left(1-\frac{1}{n}\right)^{n^2}\right]^{1/n}=\left(1-\frac{1}{n}\right)^n\xrightarrow [n\to\infty]{}e^{-1}<1 ...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    \displaystyle{\frac{1}{\ln(1+e^n)} is monotonically descending convergent to zero so \displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{\ln(1+e^n)}} is cond. convergent.

    It isn't abs. convergent since \displaystyle{\frac{1}{\ln(1+e^n)}\geq\frac{1}{\ln (2e^n)}=\frac{1}{\ln 2+n} , which is divergent.

    For the second one the root test gives \displaystyle{\left[\left(1-\frac{1}{n}\right)^{n^2}\right]^{1/n}=\left(1-\frac{1}{n}\right)^n\xrightarrow [n\to\infty]{}e^{-1}<1 ...

    Tonio
    do you use the alternating series test for the first one, I dun quite understand how it works, I understanf why is not absolutely convergent, but why is convergent without the absolute value sign?

    and can you explain why  \frac{1}{\ln 2+n} is diverges, it seems going to 0 as n goes to infinty
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    Quote Originally Posted by wopashui View Post
    do you use the alternating series test for the first one, I dun quite understand how it works, I understanf why is not absolutely convergent, but why is convergent without the absolute value sign?


    Read about Leibnitz Series.

    and can you explain why  \frac{1}{\ln 2+n} is diverges, it seems going to 0 as n goes to infinty

    The sequence does, but not the series formed with it, which is what we were talking about.

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    \displaystyle{\frac{1}{\ln(1+e^n)} is monotonically descending convergent to zero so \displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{\ln(1+e^n)}} is cond. convergent.

    It isn't abs. convergent since \displaystyle{\frac{1}{\ln(1+e^n)}\geq\frac{1}{\ln (2e^n)}=\frac{1}{\ln 2+n} , which is divergent.

    For the second one the root test gives \displaystyle{\left[\left(1-\frac{1}{n}\right)^{n^2}\right]^{1/n}=\left(1-\frac{1}{n}\right)^n\xrightarrow [n\to\infty]{}e^{-1}<1 ...

    Tonio
    so you said ] \displaystyle{\frac{1}{\ln(1+e^n)} converges to zero then you said it is diverges, so is it diverges or converges? and why is \frac{1}{\ln 2+n} divers, it seems approaching to 0 as n goes to infinty
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  6. #6
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    Quote Originally Posted by wopashui View Post
    so you said ] \displaystyle{\frac{1}{\ln(1+e^n)} converges to zero then you said it is diverges, so is it diverges or converges? and why is \frac{1}{\ln 2+n} divers, it seems approaching to 0 as n goes to infinty

    You continue to confuse infinite series with the infinite sequence it is formed with!

    One thing is a series \sum\limits^\infty_{n=1}a_n , and another one is its sequence \left\{a_n\right\}^\infty_{n=1} ...

    Tonio
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  7. #7
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    Quote Originally Posted by tonio View Post
    You continue to confuse infinite series with the infinite sequence it is formed with!

    One thing is a series \sum\limits^\infty_{n=1}a_n , and another one is its sequence \left\{a_n\right\}^\infty_{n=1} ...

    Tonio
    so, the series 1/ln2+n >1/2n which is the multiple of the harmonic series, so the series diverges, right?
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