# Thread: sum of series help..

1. ## sum of series help..

hi, just wanted some guidance on these problems.

so is this right..

using
$\displaystyle \sum\limits_{n = 0}^\infty {x^n = \frac{1} {{1 - x}}}$

$\displaystyle |x| < 1$

$\displaystyle \sum\limits_{n = 0}^\infty {(\frac{9} {{10}})^n = \frac{1} {{1 - 9/10}}}$

$\displaystyle 1/1/10 = 10$

that is the sum right?

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

$\displaystyle \sum\limits_{n = 0}^\infty {n(\frac{9} {{10}})^n }$

2. Yes, multiply your result from the first one by 9.

3. If we let the original series be S and factor, we can get

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

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

Factor out 9/10:

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

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

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

Solve for S=90

4. Originally Posted by dankelly07

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

$\displaystyle \sum\limits_{n = 0}^\infty {n(\frac{9} {{10}})^n }$
This is like $\displaystyle S=\sum_{n=0}^\infty nx^n$

Factor out x. This gives :

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

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

Therefore $\displaystyle \sum_{n=0}^\infty n\left(\tfrac 9{10}\right)^n=\tfrac 9{10} \cdot \frac{1}{(1-\tfrac 9{10})^2}=\tfrac 9{10} \cdot 100=90$

5. thanks alot, both answers help alot,

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

$\displaystyle S = \sum\limits_{n = 0}^\infty {n^2 } x^n$

factor out the x

$\displaystyle s = x\sum\limits_{n = 0}^\infty {nx^{n - 2} }$

(derivative of

$\displaystyle \sum\limits_{n = 0}^\infty {nx^{n - 1} }$)

so..

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

Close??

6. oops i think it was supposed to be 2/(1-x^3)
wasn't it?

7. Originally Posted by dankelly07
thanks alot, both answers help alot,

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

$\displaystyle S = \sum\limits_{n = 0}^\infty {n^2 } x^n$

factor out the x

$\displaystyle s = x\sum\limits_{n = 0}^\infty {nx^{n - 2} }$

(derivative of

$\displaystyle \sum\limits_{n = 0}^\infty {nx^{n - 1} }$)

so..

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

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

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

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

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

$\displaystyle \therefore\quad\boxed{\sum_{n=0}^{\infty}n^2x^n=x^ 2\left(\frac{1}{1-x}\right)''+x\cdot\left(\frac{1}{1-x}\right)'}$

8. You know from above that $\displaystyle \sum\limits_{n = 0}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$.
The derivative is $\displaystyle \sum\limits_{n = 0}^\infty {n^2 x^{n - 1} } = \frac{{x + 1}}{{\left( {1 - x} \right)^3 }}$.
Multiply by x, $\displaystyle \sum\limits_{n = 0}^\infty {n^2 x^n } = \frac{{x^2 + x}}{{\left( {1 - x} \right)^3 }}$.