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Math Help - Examine the convergence of the series

  1. #1
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    Examine the convergence of the series

    Question : Using the Ratio Test for the convergence of an infinite series, examine the convergence of the series

    \frac{1}{2} + \frac{1}{2.2^2}+\frac{1}{3.2^3}+ \frac{1}{4.2^4}+...

    Solution

    My work::::::::::::

    u_n = \sum_{n=1}^{ \infty} \ \frac{1}{n.2^n}.............Is this correct?????
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    Quote Originally Posted by zorro View Post
    Question : Using the Ratio Test for the convergence of an infinite series, examine the convergence of the series

    \frac{1}{2} + \frac{1}{2.2^2}+\frac{1}{3.2^3}+ \frac{1}{4.2^4}+...

    Solution

    My work::::::::::::

    u_n = \sum_{n=1}^{ \infty} \ \frac{1}{n.2^n}.............Is this correct?????

    u_n=\frac{1}{n2^n}\Longrightarrow \frac{u_{n+1}}{u_n}=\frac{1}{2}\cdot \frac{n}{n+1} \xrightarrow [n\to\infty]{}\frac{1}{2}<1

    Tonio
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  3. #3
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    Obviously \frac{1}{n2^{n}}\le \frac{1}{2^{n}}, so the general term of the series was also comparable to a geometric one.
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  4. #4
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    Quote Originally Posted by tonio View Post
    u_n=\frac{1}{n2^n}\Longrightarrow \frac{u_{n+1}}{u_n}=\frac{1}{2}\cdot \frac{n}{n+1} \xrightarrow [n\to\infty]{}\frac{1}{2}<1

    Tonio
    No tonio i meant the formation of the summation is that correct ??????

    I have already got the answer which u have provided........
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    Quote Originally Posted by zorro View Post
    No tonio i meant the formation of the summation is that correct ??????

    I have already got the answer which u have provided........

    If you're asking whether the series is what you wrote then the answer is yes, but I don't understand why you wrote u_n ...
    Perhaps you meant to write the n-th term of the sequence of partials sums of the series? If so then it is u_n=\sum\limits_{k=1}^n\frac{1}{k2^k}

    Tonio
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    no i wanted to know if the u_n which i have used is corect or know because only up on that the u_{n+1} would be decided that is why wnated to know if the starting is correct or no
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    Quote Originally Posted by zorro View Post
    no i wanted to know if the u_n which i have used is corect or know because only up on that the u_{n+1} would be decided that is why wnated to know if the starting is correct or no

    Ok, I don't understand what you're asking, but your u_n=\sum\limits_{n=1}^\infty\frac{1}{n2^n} is the WHOLE series...! It makes no sense.

    Tonio
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  8. #8
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    Quote Originally Posted by zorro View Post
    Question : Using the Ratio Test for the convergence of an infinite series, examine the convergence of the series

    \frac{1}{2} + \frac{1}{2.2^2}+\frac{1}{3.2^3}+ \frac{1}{4.2^4}+...

    Solution

    My work::::::::::::

    u_n = \sum_{n=1}^{ \infty} \ \frac{1}{n.2^n}.............Is this correct?????
    Learning to use "Ratio Test for the convergence" is important. But sometimes you can "eyeball" the answer by looking at the form of the series. I generally enjoy summing up such series. I am sorry for swaying from the main topic, but I want to demonstrate this method. It wont work everywhere but with experience you will know where to use it

    The 'n' factor in the denominator looks as though it x^{n-1} has been integrated. It is immediate that

    for |x| < 1,
    \frac1{1-y} = \sum_{n=1}^{ \infty} y^{n-1}
    \int_0^x \frac1{1-y}\, dy = \sum_{n=1}^{ \infty} \int_0^x y^{n-1}\, dy
    -\ln(1-y)\bigg{|}_0^x = \sum_{n=1}^{ \infty} \frac{y^n}{n}\bigg{|}_0^x
    -\ln(1-x) = \sum_{n=1}^{ \infty} \frac{x^n}{n}
    Now substitute x = \frac12 and so you even know that \sum_{n=1}^{ \infty} \ \frac{1}{n.2^n} = \ln 2
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  9. #9
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    Thanks mite
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