Question 1117305
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<pre>sum[i=1,infty] [-4*(1/3)^(n-1)]</pre> is an understandable plain text representation of 


*[tex \LARGE \sum_{i=1}^\infty\,-4\left(\frac{1}{3}\right)^{n-1}]


*[tex \LARGE -4\left(\frac{1}{3}\right)^0\ -4\left(\frac{1}{3}\right)^1\ -4\left(\frac{1}{3}\right)^2\ -4\left(\frac{1}{3}\right)^3 = -4\ -\ \frac{4}{3}\ -\ \frac{4}{9}\ -\ \frac{4}{27}\ \approx\ -5.93]


As to convergence/divergence, there are a couple of ways to look at it.


You can write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^\infty\,-4\left(\frac{1}{3}\right)^{n-1}\ =\ -4\,\sum_{i=0}^\infty\,\left(\frac{1}{3}\right)^n]


But *[tex \LARGE \sum_{i=0}^\infty\,\left(\frac{1}{3}\right)^n] is a geometric series with *[tex \LARGE r\ =\ \frac{1}{3}].  Since any geometric series with *[tex \LARGE -1\ <\ r\ <\ 1] converges, this series must converge.


Alternatively, you can use the Series Ratio Test:  A series where:


*[tex \LARGE \lim_{n\right\infty}\,\frac{a_{n+1}}{a_n}\ =\ L]


where *[tex \LARGE L\ <\ 1] converges.  Verification of the series ratio test is left as an exercise for the student.


Either way, the series converges.


The sum of a geometric series with a common ratio *[tex \Large r] is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S\ =\ \frac{1}{1-r}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4\,\sum_{i=0}^\infty\,\left(\frac{1}{3}\right)^n\ =\ -4\left(\frac{1}{1\ -\ \frac{1}{3}}\right)\ =\ -6]


You can verify the arithmetic for yourself.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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