Identify whether the series summation of 12 open parentheses 3 over 5 close parentheses to the i minus 1 power from 1 to infinity is a convergent or divergent geometric series and find the sum, if possible.

Question

sillybilly123 · Answer

It looks like you have this:

$S = \sum\limits_{i = 1}^{\infty} 12 \left( \frac{3}{5 } \right)^{i - 1} \qquad \square $

It is always well worth writing out a few terms if patterns just don't leap out at you. And if $\square $ is correct, it is this **geometric** series, with the 12 factored out.

$S = 12 \left( 1 + \left( \frac{3}{5 } \right) + \left( \frac{3}{5 } \right)^2 + \cdots \right)  $

And so from the summation of a geometric series, $S_n = a (\frac{1-r^n}{1-r})$ - you should know how to derive that formula, BTW -  you know it sums out like this:

$S_{n \to \infty} = 12 \cdot  \dfrac{1 - ({3 \over 5})^{n \to \infty}}{1 - {3 \over 5}} $

And figuring that out also confirms convergence ... as a convergent series has a finite sum

If you want to confirm convergence first before finding the sum, you might want to try the ratio test ...but not really necessary here :)