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Miscellaneous series
Note that this is the same asOriginally Posted by matsci0000
.
Ifand
, 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](http://latex.codecogs.com/png.latex?\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)}](http://latex.codecogs.com/png.latex?=\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.
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