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Math Help - An easy infinite series

  1. #1
    MHF Contributor arbolis's Avatar
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    An easy infinite series

    I must look pathetic, but I'm beginning with infinite series.
    I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
    The series is : \sum_{i=1}^{+\infty} \frac{i}{3^i}.
    I tried the ratio test, writing down a_i and a_{i+1}, dividing a_i by a_{i+1} I finally got \frac{3i}{i+1}. Taking the limit when i tends to +\infty I got 3. As 3>1, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I must look pathetic, but I'm beginning with infinite series.
    I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
    The series is : \sum_{i=1}^{+\infty} \frac{i}{3^i}.
    I tried the ratio test, writing down a_i and a_{i+1}, dividing a_i by a_{i+1} I finally got \frac{3i}{i+1}. Taking the limit when i tends to +\infty I got 3. As 3>1, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
    Root test

    \lim_{i\to\infty}\bigg(\frac{i}{3^i}\bigg)^{\frac{  1}{i}}=\lim_{i\to\infty}\frac{i^{\frac{1}{i}}}{3}=  \frac{1}{3}<1

    thus convergent
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I must look pathetic, but I'm beginning with infinite series.
    I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
    The series is : \sum_{i=1}^{+\infty} \frac{i}{3^i}.
    I tried the ratio test, writing down a_i and a_{i+1}, dividing a_i by a_{i+1} I finally got \frac{3i}{i+1}. Taking the limit when i tends to +\infty I got 3. As 3>1, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
    Also by ratio test

    \lim_{i\to\infty}\bigg|\frac{i+1}{3\cdot{3^i}}\cdo  t\frac{3^i}{i}\bigg|=\lim_{i\to\infty}\bigg|\frac{  i+1}{3i}\bigg|=\frac{1}{3}

    Remember the ratio and root test always yield the same number
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  4. #4
    Eater of Worlds
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    You can use the geometric series formula.

    \sum_{i=1}^{\infty}\frac{i}{3^{i}}=\frac{1}{2}\cdo  t\frac{1}{1-\frac{1}{3}}=\frac{3}{4}
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    Quote Originally Posted by galactus View Post
    You can use the geometric series formula.

    \sum_{i=1}^{\infty}\frac{i}{3^{i}}=\frac{1}{2}\cdo  t\frac{1}{1-\frac{1}{3}}=\frac{3}{4}
    But this is not geometric, it has an extra term in it. It happen to the the derivative of the geometric series.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I must look pathetic, but I'm beginning with infinite series.
    I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
    The series is : \sum_{i=1}^{+\infty} \frac{i}{3^i}.
    I tried the ratio test, writing down a_i and a_{i+1}, dividing a_i by a_{i+1} I finally got \frac{3i}{i+1}. Taking the limit when i tends to +\infty I got 3. As 3>1, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
    Also

    There \exists{N}\backepsilon\forall{n>N}\text{ }\frac{n}{3^n}<\frac{2^n}{3^n}=\bigg(\frac{2}{3}\b  igg)^n

    Thus convergent by comparison test

    Quote Originally Posted by arbolis View Post
    I must look pathetic, but I'm beginning with infinite series.
    I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
    The series is : \sum_{i=1}^{+\infty} \frac{i}{3^i}.
    I tried the ratio test, writing down a_i and a_{i+1}, dividing a_i by a_{i+1} I finally got \frac{3i}{i+1}. Taking the limit when i tends to +\infty I got 3. As 3>1, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
    Also
    since a_{i+1}<a_i for a finite value of i and every term is positive
    \int_1^{\infty}\frac{x}{3^x}dx=.58

    thus the integral converges by the integral test


    if you would like I can give you some practice problems and help you through them

    Quote Originally Posted by galactus View Post
    You can use the geometric series formula.

    \sum_{i=1}^{\infty}\frac{i}{3^{i}}=\frac{1}{2}\cdo  t\frac{1}{1-\frac{1}{3}}=\frac{3}{4}
    This is equal to x\cdot\frac{d}{dx}\bigg[\sum_{n=1}^{\infty}\frac{1}{3^x}\bigg]=\frac{x}{(1-x)^2}

    evaluating at 1/3 gives the correct result...which was 3/4
    Last edited by ThePerfectHacker; May 22nd 2008 at 07:43 PM.
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  7. #7
    MHF Contributor arbolis's Avatar
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    I just realized I made the ratio test but inversed! So I got 3 instead of \frac{1}{3}. I also notice that the series value is not equal to the result of the ratio test (yeah, why would it be so?).
    From Mathstud :
    Also
    since a_{i+1}<a_i for a finite value of i and every term is positive
    \int_1^{\infty}\frac{x}{3^x}dx=.58
    . Is this the value of the series? If yes, any series can be written as an improper integral and the value of the series is always equals to the improper integral?

    To
    if you would like I can give you some practice problems and help you through them
    I think it's a good idea, but I'm not sure I will do a lot of them these days (since I have an exam coming very soon and it has almost nothing to see with infinite series). But why not doing a bit? So yes.
    Finally I didn't understand well the geometric series formula. I will investigate about it.
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  8. #8
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    Quote Originally Posted by arbolis View Post
    I must look pathetic, but I'm beginning with infinite series.
    I invented to myself a very easy infinite series that I must determine its convergence. (I'm sure it converges even if I'm not able to prove it!).
    The series is : \sum_{i=1}^{+\infty} \frac{i}{3^i}.
    I tried the ratio test, writing down a_i and a_{i+1}, dividing a_i by a_{i+1} I finally got \frac{3i}{i+1}. Taking the limit when i tends to +\infty I got 3. As 3>1, the series diverges. Where is my mistake? Also, I would like to know if I can know to what number it converges.
    \sum_{i=1}^{\infty}{i}{x^i}=x \sum_{i=1}^{\infty}\frac{d}{dx}{x^i}=x \times \frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{x}{(1-x)^2}

    which all works for |x|<1, so put x=1/3, to get:

    \sum_{i=1}^{\infty}\frac{i}{3^i}=\frac{(1/3)}{(2/3)^2}=\frac{3}{4}

    RonL
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  9. #9
    MHF Contributor arbolis's Avatar
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    Thanks CaptainBlack! Nicely explained. (Sorry if I didn't quote your answer, I have some problem with the quoting part).
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  10. #10
    Eater of Worlds
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    Quote Originally Posted by Mathstud28 View Post
    Also

    There \exists{N}\backepsilon\forall{n>N}\text{ }\frac{n}{3^n}<\frac{2^n}{3^n}=\bigg(\frac{2}{3}\b  igg)^n

    Thus convergent by comparison test


    Also
    since a_{i+1}<a_i for a finite value of i and every term is positive
    \int_1^{\infty}\frac{x}{3^x}dx=.58

    thus the integral converges by the integral test


    if you would like I can give you some practice problems and help you through them


    This is equal to x\cdot\frac{d}{dx}\bigg[\sum_{n=1}^{\infty}\frac{1}{3^x}\bigg]=\frac{x}{(1-x)^2}

    evaluating at 1/3 gives the correct result...which was 3/4
    Yes, I reckon I should have explained more thoroughly. It was late and I was getting droopy.
    Sorry about that.
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  11. #11
    Super Member PaulRS's Avatar
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    G(x)=\sum_{k=0}^{\infty}{a_k\cdot{x^k}} where we want a_k=k (we'll assume x will be on the interval of convergence, that is (-1;1) )

    Now the condition can be stated as:
    a_{k+1}=a_k+1 where a_0=0

    Multiplying by x^{k+1} on both sides:
    a_{k+1}\cdot{x^{k+1}}=a_k\cdot{x^{k+1}}+x^{k+1}
    That is:
    a_{k+1}\cdot{x^{k+1}}=x\cdot{a_k\cdot{x^{k}}}+x^{k  +1}

    Now sum from k=0 to +\infty:
    \sum_{k=0}^{\infty}{a_{k+1}\cdot{x^{k+1}}}=x\cdot{  \sum_{k=0}^{\infty}{a_{k}\cdot{x^{k}}}}+\sum_{k=0}  ^{\infty}{x^{k+1}}

    Thus we have:
    G(x)-a_0=x\cdot{G(x)}+\sum_{k=1}^{\infty}{x^{k}}=x\cdot  {G(x)}+\frac{x}{1-x}

    Thus, since a_0=0:
    (1-x)\cdot{G(x)}=\frac{x}{1-x} and G(x)=\frac{x}{(1-x)^2}

    You can use this method to work with other sequences where you have a recurrence

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