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$$\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$$

It's geometric, since the common ratio $r$ appears to be $\frac{-3}{4}$, but this is where I get stuck. I think I need to do this: let $f(x) = \frac{(-3)^{x-1}}{4^x}$.

$$\lim\limits_{x \to \infty}\frac{(-3)^{x-1}}{4^x}$$

Is this how I handle this exercise? I still cannot seem to get the answer $\frac{1}{7}$

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$$(-3)^{-1} \sum_{i=1}^\infty \left( -\frac{3}{4} \right)^n = (-3)^{-1} \left[ \sum_{i=0}^\infty \left( -\frac{3}{4} \right)^n -1 \right]$$ –  Siminore Jul 3 '12 at 11:59
    
Dear Jonathan, You should label the indices in your series so that they are either both $i$ or both $n$. (At the moment you have a mixture of the two.) Regards, –  Matt E Jul 3 '12 at 16:52
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4 Answers

up vote 3 down vote accepted

A geometric series is convergent if the $|r|<1$ where $r$ is the common ratio.

Let $S_n=\sum_{i=0}^n (-3/4)^i$ then $$S_n=\frac{(-3/4)^{n+1}-1}{(-3/4)-1}$$ Now take $n\rightarrow \infty$ then $$S_n\rightarrow \frac{0-1}{(-3/4)-1}=4/7$$ because $|-3/4|<1$ and so $(-3/4)^n\rightarrow 0$. Now note that your sum is $$\mbox{lim }\sum_{i=1}^{n+1}\frac{(-3)^{i-1}}{4^{i}}=\mbox{lim }\frac{1}{4}\sum_{i=1}^{n+1}\frac{(-3)^{i-1}}{4^{i-1}}=1/4.\mbox{lim }S_n=1/7$$

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But why is the sum 1/4? The method my class and book are using is a/1-r. –  Jonathan Dewein Jul 3 '12 at 12:20
    
I did not say the sum is 1/4, note that i have taken 1/4 out to get $S_n$, anyway i edited the answer to make it clearer. If you want to use that formula then $$\sum_{i=1}^\infty\frac{(-3)^{i-1}}{4^i}=(-3)^{-1}\sum_{i=1}^\infty \left(\frac{-3}{4}\right)^i=(-3)^{-1}\frac{-3/4}{1-(-3/4)}=1/7$$ –  pritam Jul 3 '12 at 12:37
    
I see. I was factoring out a 4 when it should have been 1/4. Thank you. –  Jonathan Dewein Jul 3 '12 at 15:51
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If $\,a, ar, ar^2,...\,$ is a geometric series with $\,|r|<1\,$ ,then $$\sum_{n=0}^\infty ar^n=\lim_{n\to\infty} ar^n=\lim_{n=0}\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}$$since $\,r^n\xrightarrow [n\to\infty]{} 0\Longleftrightarrow |r|<1\,$ , and thus

$$\sum_{n=1}^\infty\frac{(-3)^{n-1}}{4^n}=\frac{1}{4}\sum_{n=0}^\infty \left(-\frac{3}{4}\right)^n=\frac{1}{4}\frac{1}{1-\left(-\left(\frac{3}{4}\right)\right)}=\frac{1}{4}\frac{1}{\frac{7}{4}}=\frac{1}{7}$$

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Let $q = \frac{-3}{4}$, $a_n = \frac{(-3)^{n-1}}{4^n}$, $b_0 = 0$, $b_n = b_{n-1} + a_n$.

$a_n = -\frac{1}{3} (\frac{-3}{4})^n = -\frac{1}{3} q^n$, hence $b_n = -\frac{1}{3} c_n$, where $c_0 = 0$, $c_n = c_{n-1} + q^n$.

The $c_n$ limit is equal to $q + q^2 + q^3 + \ldots = \frac{q}{1-q} = \frac{\frac{-3}{4}}{1-\frac{-3}{4}} = \frac{-3}{7}$, thus $b_n$ limit is equal to $\frac{1}{7}$.

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$By the given series have:

$a_{n}=\frac{(-3)^{n-1}}{4^{n}}$, $a_{n+1}=\frac{(-3)^{n}}{4^{n+1}}$

By the criterion of Dalamber have:

$A=\lim\frac{a_{n+1}}{a_{n}}=\frac{3}{4}<1$

Under this criterion we have that A<1 conclude that given series is convergent.

Since the given series is convergent exist sum of this series. Mark wis $S_{n}$ sum of this series, it is $S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$.

Hance we

$S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$

$S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty \frac{(-3)^{n}}{4^n}$

$\frac{(-3)^{n}}{4^n}$ is it a geometric series. Now find sum this series. Find the sum must assign a_{1} dhe q.

$a_{1}=-\frac{3}{4}$, $q=-\frac{3}{4}$.

Sum accounst

$S_{n}=\frac {a_{1}(1-q^{n})}{1-q}=-\frac{3}{7}{[1-\frac{3^{n}}{4^{n}}]}$

$\lim S_{n}=-\frac{3}{7}$

Theres definitely have

$S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty\frac{(-3)^{n}}{4^{n}}$

$S_{n}=(-\frac{1}{3})(-\frac{3}{7})$

$S_{n}=\frac{1}{7}$

Conlude the: Given series is the convergent and its sum $\frac{1}{7}$.$

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Please do use $$ to wrap mathematical formulae, not all the text. –  DonAntonio Jul 3 '12 at 14:08
    
I tried to edit your post but didn't succeed to guess all your ideas. You must be way more careful when you write down mathematics, and also you must learn how to use LaTeX in this site (and enhancing your english won't hurt, either). Nice effort for a beginner, though. –  DonAntonio Jul 3 '12 at 14:22
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