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Math Help - sum of series help..

  1. #1
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    sum of series help..

    hi, just wanted some guidance on these problems.

    so is this right..

    using
    <br />
\sum\limits_{n = 0}^\infty  {x^n  = \frac{1}<br />
{{1 - x}}} <br />

    <br />
|x| < 1<br />

    <br />
\sum\limits_{n = 0}^\infty  {(\frac{9}<br />
{{10}})^n  = \frac{1}<br />
{{1 - 9/10}}} <br />

    <br />
1/1/10 = 10<br />

    that is the sum right?

    also how would I solve for, i'm guessing there is also a trick for this..

    <br />
\sum\limits_{n = 0}^\infty  {n(\frac{9}<br />
{{10}})^n } <br />
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  2. #2
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    Yes, multiply your result from the first one by 9.
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  3. #3
    Eater of Worlds
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    If we let the original series be S and factor, we can get

    S=\frac{9}{10}(S+9)+\frac{9}{10} and solve for S.

    S=1\cdot \frac{9}{10}+2\cdot \frac{81}{100}+3\cdot \frac{729}{1000}+....

    Factor out 9/10:

    S=\frac{9}{10}\left(1+\underbrace{2\cdot \frac{9}{10}+3\cdot \frac{81}{100}+4\cdot \frac{729}{1000}+...}_{\text{S+9}}\right)

    But what I have underbraced is \sum_{n=1}^{\infty}(n+1)(\frac{9}{10})^{n}=\sum_{n  =1}^{\infty}n(\frac{9}{10})^{n}+\sum_{n=1}^{\infty  }(\frac{9}{10})^{n}=S+9

    So what we get is S=\frac{9}{10}(S+9)+\frac{9}{10}

    Solve for S=90
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  4. #4
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    Quote Originally Posted by dankelly07 View Post

    also how would I solve for, i'm guessing there is also a trick for this..

    <br />
\sum\limits_{n = 0}^\infty  {n(\frac{9}<br />
{{10}})^n } <br />
    This is like S=\sum_{n=0}^\infty nx^n

    Factor out x. This gives :

    S=x \sum_{n=0}^\infty n x^{n-1}
    You can see that \sum_{n=0}^\infty n x^{n-1} is the derivative of \sum_{n=0}^\infty x^n, which is \frac{1}{1-x} for |x|<1.

    Thus S=x \cdot \left(\frac{1}{1-x}\right)'=x \cdot \frac{1}{(1-x)^2}

    Therefore \sum_{n=0}^\infty n\left(\tfrac 9{10}\right)^n=\tfrac 9{10} \cdot \frac{1}{(1-\tfrac 9{10})^2}=\tfrac 9{10} \cdot 100=90
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  5. #5
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    thanks alot, both answers help alot,

    just to make sure i've got it can you just double check I'm doing this right?

    <br />
S = \sum\limits_{n = 0}^\infty  {n^2 } x^n <br />


    factor out the x

    <br />
s = x\sum\limits_{n = 0}^\infty  {nx^{n - 2} } <br />

    (derivative of

    <br />
\sum\limits_{n = 0}^\infty  {nx^{n - 1} } <br />
)

    so..


    <br />
\begin{gathered}<br />
  s = x.\left( {\frac{1}<br />
{{1 - x}}} \right)'' \hfill \\<br />
= x.\left( {\frac{1}<br />
{{1 - x^3 }}} \right) \hfill \\<br />
  \sum\limits_{n = 0}^\infty  {n^2 \left( {\frac{9}<br />
{{10}}} \right)^n  = \frac{9}<br />
{{10}}.\frac{1}<br />
{{(1 - \frac{9}<br />
{{10}})^3 }} = 900}  \hfill \\ <br />
\end{gathered} <br />

    Close??
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  6. #6
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    oops i think it was supposed to be 2/(1-x^3)
    wasn't it?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dankelly07 View Post
    thanks alot, both answers help alot,

    just to make sure i've got it can you just double check I'm doing this right?

    <br />
S = \sum\limits_{n = 0}^\infty {n^2 } x^n <br />


    factor out the x

    <br />
s = x\sum\limits_{n = 0}^\infty {nx^{n - 2} } <br />

    (derivative of

    <br />
\sum\limits_{n = 0}^\infty {nx^{n - 1} } <br />
)

    so..


    <br />
\begin{gathered}<br />
s = x.\left( {\frac{1}<br />
{{1 - x}}} \right)'' \hfill \\<br />
= x.\left( {\frac{1}<br />
{{1 - x^3 }}} \right) \hfill \\<br />
\sum\limits_{n = 0}^\infty {n^2 \left( {\frac{9}<br />
{{10}}} \right)^n = \frac{9}<br />
{{10}}.\frac{1}<br />
{{(1 - \frac{9}<br />
{{10}})^3 }} = 900} \hfill \\ <br />
\end{gathered} <br />

    Close??
    \sum_{n=2}^{\infty}n^2x^n=x^2\cdot\left(\sum_{n=0}  ^{\infty}x^n\right)''=x^2\cdot\left(\frac{1}{1-x}\right)'' if that is what you are trying to say that is wrong

    \left(\sum_{n=0}^{\infty}x^n\right)''=\sum_{n=2}^{  \infty}n(n-1)x^n

    \underbrace{=}_{\cdot{x^2}}\sum_{n=0}^{\infty}n^2x  ^n-\sum_{n=0}^{\infty}nx^n

    =\sum_{n=0}^{\infty}n^2x^n-x\cdot\left(\frac{1}{1-x}\right)'

    \therefore\quad\boxed{\sum_{n=0}^{\infty}n^2x^n=x^  2\left(\frac{1}{1-x}\right)''+x\cdot\left(\frac{1}{1-x}\right)'}
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  8. #8
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    You know from above that \sum\limits_{n = 0}^\infty  {nx^n }  = \frac{x}{{\left( {1 - x} \right)^2 }}.
    The derivative is \sum\limits_{n = 0}^\infty  {n^2 x^{n - 1} }  = \frac{{x + 1}}{{\left( {1 - x} \right)^3 }}.
    Multiply by x, \sum\limits_{n = 0}^\infty  {n^2 x^n }  = \frac{{x^2  + x}}{{\left( {1 - x} \right)^3 }}.
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