Sum to n of series

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August 16th 2010, 11:13 PM
arze

Sum to n of series

If $S_n=\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$ , $(|a|\neq 1)$ by considering $(1-a)S_n$ , show that
$S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
State the set of values of a for which $S_n$ approaches a limit as $n \rightarrow\infty$ and find the sum to infinity of the series for these values.

Attempt:
$\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$
$=\displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^n_{r=0}a^r\right)$
$=\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^r}{1-a}\right)$
$=\displaystyle \sum^n_{r=0}\frac{a^r-a^{2r}}{1-a}$
Am I correct so far? and how to continue because I don't see a way of summing that?
Thanks!
August 17th 2010, 01:25 AM
Vlasev

This is almost correct. Here is the correct version:

$\displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^r_{k=0}a^k\right)$

$=\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^{r+1}}{1-a}\right)$

In the question you are told to consider $(1-a)S_n$ . Why don't you try that!
August 17th 2010, 01:59 AM
arze

$(1-a)S_n=S_n-aS_n$
$=\displaystyle \sum^n_{r=0}a^r\left(\frac{(1-a^{r+1})(1-a)}{1-a}\right)$
$=\displaystyle \sum^n_{r=0}a^r(1-a^{r+1})$
$\displaystyle \sum^n_{r=0}a^r-\sum^n_{r=0}a^{2r+1}$
$=\frac{1-a^{n+1}}{1-a}-\frac{a-a^{2n+2}}{1-a^2}$

$S_n=\frac{1-a^{n+1}}{(1-a)^2}-\frac{a-a^{2n+2}}{(1-a^2)(1-a)}$
Which isn't the given one and I can't see where I've gone wrong (Worried)
Thanks!
August 17th 2010, 02:11 AM
Sudharaka

Quote:

Originally Posted by arze

If $S_n=\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$ , $(|a|\neq 1)$ by considering $(1-a)S_n$ , show that
$S_n=\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
State the set of values of a for which $S_n$ approaches a limit as $n \rightarrow\infty$ and find the sum to infinity of the series for these values.

Attempt:
$\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$
$=\displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^n_{r=0}a^r\right)$
$=\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^r}{1-a}\right)$
$=\displaystyle \sum^n_{r=0}\frac{a^r-a^{2r}}{1-a}$
Am I correct so far? and how to continue because I don't see a way of summing that?
Thanks!

Dear arze,

When n=1,

$S_1=\displaystyle\sum_{r=0}^{1}{a^r(1+a+a^2+.....+ a^r)}=a^0+a(1+a)$

But if you consider the expression, $\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$ ......

When n=1,

$\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$

$=\frac{1-a^{4}}{(1-a)^3}-\frac{a^{2}(1+a^{2})}{(1-a)^2}$

$=\frac{1+a^2}{1-a^2}\left(\frac{1-a^2}{1-a}-a^2\right)$

$=\frac{1+a^2}{1-a^2}\left(1+a-a^2\right)$

Therefore, $S_1\neq\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
August 17th 2010, 02:26 AM
Vlasev

Yea, that's not the correct identity. The correct identity is

[tex]\displaystyle \displaystyle S_n = \frac{(1-a^{n+1})(1-a^{n+2})}{(1-a)^2(1+a)}[/Math]
August 17th 2010, 03:57 AM
arze

Quote:

Originally Posted by Vlasev

Yea, that's not the correct identity. The correct identity is

[tex]\displaystyle \displaystyle S_n = \frac{(1-a^{n+1})(1-a^{n+2})}{(1-a)^2(1+a)}[/Math]

But that doesn't give me the result I'm supposed to get either
$S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
August 17th 2010, 04:57 AM
Vlasev

Quote:

Originally Posted by arze

But that doesn't give me the result I'm supposed to get either
$S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$

Because what you are asked to get is (in this form) false! Unless you typed something incorrectly in your question.
August 17th 2010, 05:36 AM
arze

hmmm, i checked the book again and i did not type anything wrong. Well, thank you very much!
August 17th 2010, 06:55 AM
Archie Meade

$\displaystyle\ S_n=\sum_{r=0}^na^r\left(1+a+a^2+....+a^r\right)$

$\displaystyle\ aS_n=\sum_{r=0}^na^{r+1}\left(1+a+a^2+....+a^r\rig ht)$

$\displaystyle\ (1-a)S_n=\sum_{r=0}^n\left(a^r+a^{r+1}+a^{r+2}+....+a ^{2r}\right)-\sum_{r=0}^n\left(a^{r+1}+a^{r+2}+....+a^{2r+1}\ri ght)$

$\displaystyle\ =\sum_{r=0}^na^r-\sum_{r=0}^na^{2r+1}$

The first sum has $n+1$ terms, the first term is 1 and the common ratio is $a$ .
The second sum has $(2n+2)/2 =n+1$ terms, the first term is $a$ and the common ratio is $a^2$ .

$\displaystyle\ (1-a)S_n=\frac{1-a^{n+1}}{1-a}-a\left(\frac{1-a^{2(n+1)}}{1-a^2}\right)$

$\displaystyle\ S_n=\frac{1-a^{n+1}}{(1-a)^2}-a\left(\frac{1-a^{2n+2}}{(1-a)^2(1+a)}\right)$

$\displaystyle\huge\ =\frac{(1+a)\left(1-a^{n+1}\right)-a\left(1-a^{2n+2}\right)}{(1-a)^2(1+a)}$

$\displaystyle\huge\ =\frac{(1+a)\left(1-a^{n+1}\right)-a\left(1-a^{n+1}\right)\left(1+a^{n+1}\right)}{(1-a)^2(1+a)}$

$\displaystyle\huge\ =\frac{\left(1-a^{n+1}\right)\left[1+a-a\left(1+a^{n+1}\right)\right]}{(1-a)^2(1+a)}$

$\displaystyle\huge\ =\frac{\left(1-a^{n+2}\right)\left(1-a^{n+1}\right)}{(1-a)^2(1+a)}$

so I'd have to agree with Vlasev.