# Working with exponent on series

Hi have this sequence:

$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$

I understand that this is a Geometric series so this is what I've made to get the sum. $$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$ $$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4})}^n$$

So $a= (-1)^n\cdot 3^{-2}$ and $r=\frac{3}{4}$ and the sum is given by $$(-1)^n\cdot 3^{-2}\cdot \frac{1}{1-\frac{3}{4}}$$

Solving this I'm getting the result as $\frac{4}{9}$ witch I know Is incorrect because WolframAlpha is giving me another result.

So were am I making the mistake?

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1
The ratio of your geometric series is $r=-3/4$, not $3/4$. The term $a$ must be a constant, it can not depend on $n$. – julien Jan 26 at 16:30
Please use markdown rather than LaTeX for text formatting ("Geometric series"). There's fairly extensive help on markdown syntax if you click the little ? above the right side of the text-entry box. – Jonathan Christensen Jan 26 at 16:32
you cannot exract $(-1)^n$ because the variable $n$ is involved. Instead you must leave it at the quotient $\frac 34$... – Gottfried Helms Jan 26 at 16:33

## 3 Answers

The objective here is to transform your sum into a sum of the form:

$$\sum_{n=1}^\infty ar^{n-1}$$

$$\text{Transformation: }\quad\quad\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-3)^{n-1}}{4^{n-1}} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\left(\frac{-3}{4}\right)^{n-1}$$

Hence $a = -\dfrac{1}{12}$ and $r = -\dfrac{3}{4}.\quad$ Now use the fact that

$$\sum_{n=1}^\infty ar^{n-1} = \dfrac{a}{1 - r} = -\left(\frac{1}{12}\right)\cdot \left(\frac{1}{1 - (-\frac{3}{4})}\right)$$

Simpilfy, and then you are done!

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There is something wrong about the second term of your equalities. – julien Jan 26 at 16:39
Not correct, because $3^{n-2}=3^{n-1}3^{-1}=\frac{1}{3}3^{n-1}$ and not $3^{n-2}=3*3^{n-1}=$ In addition, wrong handling of the term $(-1)^n$ – Tomas Jan 26 at 16:42
@amWhy Thanks. Since $n=1$ you transform the $r^{n-1}$. But if $n=4$? It's still valid what you have done? – Favolas Jan 26 at 16:44
Thanks, Thomas! You beat me to it... – amWhy Jan 26 at 16:47

1
@Babak Hahaha...you're certainly not under zero! ;-) – amWhy Jan 26 at 19:56
show 5 more comments

$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}=\sum\limits_{n=1}^\infty (-1)^n\frac{1}{9}\left(\frac{3}{4}\right)^n=\frac{1}{9}\sum\limits_{n=1}^\infty \left(-\frac{3}{4}\right)^n= \frac{1}{9}\left(\frac{1}{1+3/4}-1\right)$$

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No, it sould be 1/9 instead 9 – Tomas Jan 26 at 16:43
Yes i miss that. Thanks – Adi Dani Jan 26 at 16:49

You have

$$\begin{align} \sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} &= \sum_{n=1}^{\infty} \frac{(-1)(-1)^{n-1}\frac{1}{3}3^{n-1}}{4\cdot 4^{n-1}} \\ &= \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-1)^{n-1}3^{n-1}}{4^{n-1}}\\ &= \sum_{n=1}^{\infty} \frac{-1}{12}\left(\frac{-3}{4}\right)^{n-1} \end{align}$$ So you have $a = \frac{-1}{12}$ and $r = \frac{-3}{4}$.

Note the key thing here that both $a$ and $r$ are constants/numbers. They do not depend on $n$. The idea is that you rewrite your series so that it is of exactly the form $$\sum_{n=1}^{\infty} a r^{n-1}$$ where again $a$ and $r$ are constants/numbers.

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