# sum of series help..

• Jul 28th 2008, 09:00 AM
dankelly07
sum of series help..
hi, just wanted some guidance on these problems.

so is this right..

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

$
|x| < 1
$

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

$
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..

$
\sum\limits_{n = 0}^\infty {n(\frac{9}
{{10}})^n }
$
• Jul 28th 2008, 09:49 AM
galactus
Yes, multiply your result from the first one by 9.
• Jul 28th 2008, 10:08 AM
galactus
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
• Jul 28th 2008, 10:13 AM
Moo
Quote:

Originally Posted by dankelly07

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

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

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 $\sum_{n=0}^\infty n\left(\tfrac 9{10}\right)^n=\tfrac 9{10} \cdot \frac{1}{(1-\tfrac 9{10})^2}=\tfrac 9{10} \cdot 100=90$
• Jul 28th 2008, 11:04 AM
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?

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

factor out the x

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

(derivative of

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

so..

$
\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??
• Jul 28th 2008, 12:12 PM
dankelly07
oops i think it was supposed to be 2/(1-x^3)
wasn't it?
• Jul 28th 2008, 12:59 PM
Mathstud28
Quote:

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?

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

factor out the x

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

(derivative of

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

so..

$
\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??

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

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

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

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

$\therefore\quad\boxed{\sum_{n=0}^{\infty}n^2x^n=x^ 2\left(\frac{1}{1-x}\right)''+x\cdot\left(\frac{1}{1-x}\right)'}$
• Jul 28th 2008, 02:04 PM
Plato
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, $\sum\limits_{n = 0}^\infty {n^2 x^n } = \frac{{x^2 + x}}{{\left( {1 - x} \right)^3 }}$.