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Using maclurin series to find sums to infinity? Watch

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    Hey, I'm trying to use the maclurin series for \frac{1}{1-x} to try and find...

    1)\displaystyle\sum_{n=0}^\infty\f  rac{1}{3^n}

    2)\displaystyle\sum_{n=0}^\infty\f  rac{n}{3^n}

    3)\displaystyle\sum_{n=0}^\infty\f  rac{n^2}{3^n}

    I've found the maclurin series for \frac{1}{1-x} is 1+x+x^2+x^3+....

    The question also hints to differentiate term-by-term twice, but I'm not seeing the relationship.
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    so you know how to calculate  \sum_{r=0}^{\infty} x^r . Can you think of a suitable substitution to turn x^r into  \frac{1}{3^r} ?
    If you differentiate term by term, what do you get? As a consequence, you know how to work out that sum as well.
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    ah ok, thanks, i knew I had to use x=1/3 for part i, because of the standard a/1-r formula. Just wasn't seeing the link.

    Now for part ii do I have to use the fact that...

    \displaystyle\sum_{i=0}^\infty x^r = \displaystyle\sum_{i=1}^\infty rx^{r-1}

    and also differentiating the geometric progression gives me \frac{1}{(1-x)^2}?
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    (Original post by DJB MASTER)
    Now for part ii do I have to use the fact that...

    \displaystyle\sum_{i=0}^\infty x^r = \displaystyle\sum_{i=1}^\infty rx^{r-1}
    I presume you mean the derivative of the LHS is the RHS?
    

and also differentiating the geometric progression gives me [latex]\frac{1}{(1-x)^2}?
    Yes. You can then make a substitution to get part ii

    For part iii you won't quite get r^2x^r, but you'll get something similar, so you can add something on to get r^2x^r.
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    yeh, i meant the derivativee, sry about that. I got part ii working ok.

    I'm stuck on the final part of iii...

    I say the 2nd derivative of...

    \displaystyle\sum_{i=0}^\infty x^r = \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2}

    I expand and subtract the summations...

    \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2} = \displaystyle\sum_{i=2}^\infty r^2x^{r-2} -\displaystyle\sum_{i=2}^\infty rx^{r-2}

    \displaystyle\sum_{i=2}^\infty r^2x^{r-2} = \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2} +\displaystyle\sum_{i=2}^ \infty rx^{r-2}

    \displaystyle\sum_{i=2}^\infty r^2x^r = \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2}x^2 +\displaystyle\sum_{i=2}^ \infty rx^{r-2}x^2

    Is my working correct? I can see that the middle term is my 2nd derivativee of the geometric progression (2/(1-x)^3) multiplied by x^2, where x = 1/3. It's the last summation term that has me confused.

    Is this just simply the answer I got for ii?
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    (Original post by DJB MASTER)
    yeh, i meant the derivativee, sry about that. I got part ii working ok.

    I'm stuck on the final part of iii...

    I say the 2nd derivative of...

    \displaystyle\sum_{i=0}^\infty x^r = \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2}

    I expand and subtract the summations...

    \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2} = \displaystyle\sum_{i=2}^\infty r^2x^{r-2} -\displaystyle\sum_{i=2}^\infty rx^{r-2}

    \displaystyle\sum_{i=2}^\infty r^2x^{r-2} = \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2} +\displaystyle\sum_{i=2}^ \infty rx^{r-2}

    \displaystyle\sum_{i=2}^\infty r^2x^r = \displaystyle\sum_{i=2}^\infty r(r-1)x^{r-2}x^2 +\displaystyle\sum_{i=2}^ \infty rx^{r-2}x^2

    Is my working correct? I can see that the middle term is my 2nd derivativee of the geometric progression (2/(1-x)^3) multiplied by x^2, where x = 1/3. It's the last summation term that has me confused.

    Is this just simply the answer I got for ii?
    Yes.
    You may need to be careful with the starting index of the sums. You have it correct here, but  \displaystyle\sum_{i=2}^ \infty rx^r \neq \sum_{i=0}^\infty rx^r . In order to get the other side, just add on a few numbers to the beginning (you know what  \sum_{r=0}^1 rx^r is, right?)
 
 
 
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