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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 $\displaystyle \sum_{k=0}^{\infty}\left[\sum_{j=0}^{k}a^{j}\cdot r^k\right]$.Originally Posted by matsci0000
If $\displaystyle \left|a\right|<1$ and $\displaystyle \left|r\right|<1$, then it follows that
$\displaystyle \sum_{k=0}^{\infty}\left[\sum_{j=0}^{k}a^{j}\cdot r^k\right]=\sum_{k=0}^{\infty}\left(\frac{1-a^{k+1}}{1-a}\right)r^k=\frac{1}{1-a}\sum_{k=0}^{\infty}r^k-a\left(ar\right)^k$ $\displaystyle =\frac{1}{1-a}\left[\frac{1}{1-r}-\frac{a}{1-ar}\right]=\frac{1}{1-a}\left[\frac{1-ar-a+ar}{\left(1-r\right)\left(1-ar\right)}\right]=\frac{1}{\left(1-r\right)\left(1-ar\right)}$
I imposed those conditions in order to evaluate the [geometric] series. Otherwise you can't [It diverges]!How can you make this assertion though? Is this assumed somehow? The OP didn't include any restrictions on a or r.
Hello, matsci0000!
Multiply by $\displaystyle (1-a)\!:$Find the sum: .$\displaystyle S \;=\;1+(1+a)r+(1+a+a^2)r^2+(1+a+a^2+a^3)r^3 + \hdots$
. . $\displaystyle (1-a)S \;=\;(1-a) + (1-a^2)r + (1-a^3)r^2 + (1-a^4)r^3 + \hdots $
. . $\displaystyle (1-a)S \;=\;(1 - a) + (r - a^2r) + (r^2 - a^3r^2) + (r^3 - a^4r^3) + \hdots$
. . $\displaystyle (1-a)S \;=\;\underbrace{(1 + r + r^2 + r^3 + \hdots)} - a\underbrace{(1 + ar + a^2r^2 + a^3r^3+ \hdots)} $
. . . . . . . . . . . . . . . . . . . $\displaystyle ^{\nwarrow\text{ geometric series }\nearrow} $
. . $\displaystyle (1-a)S \;=\;\frac{1}{1-r} - a\left(\frac{1}{1-ar}\right) $
. . $\displaystyle (1-a)S \;=\;\frac{1-a}{(1-r)(1-ar)} $
Therefore: .$\displaystyle S \;=\;\frac{1}{(1-r)(1-ar)}$
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