1. ## Sum

Just can't get this right.
I have -2+4/3-8/9+16/27... and want to write the sum by useing sigma.
I guess n is infinity, but after that it stopes for me. Only thing i can see is that i can multiply -2*(-2/3) to get 4/3 and so on. I tryed google to find somewhere i could learn how to think to make the formula but could not find much that help me.
Anyone can help me?

2. ## Re: Summation

Originally Posted by Mimmi
Just can't get this right.
I have -2+4/3-8/9+16/27... and want to write the sum by useing sigma.
I guess n is infinity, but after that it stopes for me. Only thing i can see is that i can multiply -2*(-2/3) to get 4/3 and so on. I tryed google to find somewhere i could learn how to think to make the formula but could not find much that help me.
Anyone can help me?
\displaystyle \begin{align*} -2 + \frac{4}{3} - \frac{8}{9} + \frac{16}{27} \pm \dots &= -\frac{2}{1} + \frac{4}{3} - \frac{8}{9} + \frac{16}{27} \pm \dots \\ &= -\frac{2^1}{3^0} + \frac{2^2}{3^1} - \frac{2^3}{3^2} + \frac{2^4}{3^3} \pm \dots \\ &= \frac{(-1)^1 \cdot 2^1}{3^0} + \frac{(-1)^2 \cdot 2^2}{3^1} + \frac{(-1)^3 \cdot 2^3}{3^2} + \frac{(-1)^4 \cdot 2^4}{3^3} + \dots \\ &= \frac{(-2)^1}{3^0} + \frac{(-2)^2}{3^1} + \frac{(-2)^3}{3^2} + \frac{(-2)^4}{3^3} + \dots \\ &= \frac{(-2)^{0 + 1}}{3^0} + \frac{(-2)^{1 + 1}}{3^1} + \frac{(-2)^{2 + 1}}{3^2} + \frac{(-2)^{3 + 1}}{3^3} + \dots \\ &= \sum_{n = 0}^{\infty}\frac{(-2)^{n + 1}}{3^n} \end{align*}

3. ## Re: Summation

Originally Posted by Mimmi
Just can't get this right.
I have -2+4/3-8/9+16/27... and want to write the sum by useing sigma.
I guess n is infinity, but after that it stopes for me. Only thing i can see is that i can multiply -2*(-2/3) to get 4/3 and so on. I tryed google to find somewhere i could learn how to think to make the formula but could not find much that help me.
Anyone can help me?
Try: $\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k 2^k }}{{3^{k - 1} }}}$

4. ## Re: Summation

I think it looks like nth term multiplied by (n+1)th term.

5. ## Re: Summation

Originally Posted by arangu1508
I think it looks like nth term multiplied by (n+1)th term.
Wrong. You can get each term by multiplying the previous one by $\displaystyle -\frac{2}{3}$...

6. ## Re: Summation

Thx! I actually understand how to do this now.