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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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    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,2012-07-03

4 Answers 4

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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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    I see. I was factoring out a 4 when it should have been 1/4. Thank you.2012-07-03
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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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    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.2012-07-03