Miscellaneous series

**matsci0000** · August 22nd 2009, 10:02 PM

Find the sum of the infinite series
1+(1+a)r+(1+a+a^2)r^2+(1+a+a^2+a^3)r^3............ .............

**Chris L T521** · August 22nd 2009, 11:08 PM

Originally Posted by matsci0000

Find the sum of the infinite series
1+(1+a)r+(1+a+a^2)r^2+(1+a+a^2+a^3)r^3............ .............

Note that this is the same as $\sum_{k=0}^{\infty}\left[\sum_{j=0}^{k}a^{j}\cdot r^k\right]$ .

If $\left|a\right|<1$ and $\left|r\right|<1$ , then it follows that

$\sum_{k=0}^{\infty}\left[\sum_{j=0}^{k}a^{j}\cdot r^k\right]=\sum_{k=0}^{\infty}\left(\frac{1-a^{k+1}}{1-a}\right)r^k=\frac{1}{1-a}\sum_{k=0}^{\infty}r^k-a\left(ar\right)^k$ $=\frac{1}{1-a}\left[\frac{1}{1-r}-\frac{a}{1-ar}\right]=\frac{1}{1-a}\left[\frac{1-ar-a+ar}{\left(1-r\right)\left(1-ar\right)}\right]=\frac{1}{\left(1-r\right)\left(1-ar\right)}$

**QM deFuturo** · August 22nd 2009, 11:15 PM

Originally Posted by Chris L T521

If $\left|a\right|<1$ and $\left|r\right|<1$ , then it follows that

How can you make this assertion though? Is this assumed somehow? The OP didn't include any restrictions on a or r.

**Chris L T521** · August 22nd 2009, 11:33 PM

Originally Posted by QM deFuturo

How can you make this assertion though? Is this assumed somehow? The OP didn't include any restrictions on a or r.

I imposed those conditions in order to evaluate the [geometric] series. Otherwise you can't [It diverges]!

**Soroban** · August 23rd 2009, 10:01 AM

Hello, matsci0000!

Find the sum: . $S \;=\;1+(1+a)r+(1+a+a^2)r^2+(1+a+a^2+a^3)r^3 + \hdots$

Multiply by $(1-a)\!:$

. . $(1-a)S \;=\;(1-a) + (1-a^2)r + (1-a^3)r^2 + (1-a^4)r^3 + \hdots$

. . $(1-a)S \;=\;(1 - a) + (r - a^2r) + (r^2 - a^3r^2) + (r^3 - a^4r^3) + \hdots$

. . $(1-a)S \;=\;\underbrace{(1 + r + r^2 + r^3 + \hdots)} - a\underbrace{(1 + ar + a^2r^2 + a^3r^3+ \hdots)}$
. . . . . . . . . . . . . . . . . . . $^{\nwarrow\text{ geometric series }\nearrow}$

. . $(1-a)S \;=\;\frac{1}{1-r} - a\left(\frac{1}{1-ar}\right)$

. . $(1-a)S \;=\;\frac{1-a}{(1-r)(1-ar)}$

Therefore: . $S \;=\;\frac{1}{(1-r)(1-ar)}$

Math Help - Miscellaneous series

LinkBack

Thread Tools

Display

Miscellaneous series

Similar Math Help Forum Discussions

Miscellaneous questions.

miscellaneous questions

A question on miscellaneous series

Miscellaneous series

[SOLVED] Miscellaneous 6 OCR

Search Tags