Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $n$. – 1015 Jan 26 '13 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 '13 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 '13 at 16:33
up vote 4 down vote accepted

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!

share|cite|improve this answer
There is something wrong about the second term of your equalities. – 1015 Jan 26 '13 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 '13 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 '13 at 16:44
Thanks, Thomas! You beat me to it... – amWhy Jan 26 '13 at 16:47
@Babak're certainly not under zero! ;-) – amWhy Jan 26 '13 at 19:56

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

share|cite|improve this answer
No, it sould be 1/9 instead 9 – Tomas Jan 26 '13 at 16:43
Yes i miss that. Thanks – Adi Dani Jan 26 '13 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.