## anonymous 4 years ago please help!! How do you solve this sigma notation with an infinite series?

1. anonymous

$\sum_{n=1}^{\infty} 1/3^n$ ---please help!!

2. anonymous

This is the sum of a geometric sequence: $\sum_{n=1}^\infty x^n = \frac{x}{1-x}$ so long as $|x| < 1$ What is x in this case?

3. anonymous

that's the thing! in the question, there is only a fraction (1/3) and n being an exponent of the denominator. it's not like the usual notation. the answer is suppose to be 1/2 but i dont know why.

4. anonymous

$\left(\frac{1}{3}\right)^n = \frac{1}{3^n}$

5. anonymous

it is the case of convergence?jemurray

6. anonymous

does that mean that there is no x?

7. anonymous

It means x=1/3. What about convergence, tanusingh?

8. anonymous

like does this infinte series converge as n tends to infinitey

9. anonymous

$\sum_{n=1}^{\infty}1/3^{n} = (1/3)/(1-(1/3))<1/2$

10. anonymous

u can also use the formula a^n+1-1/a-1

11. anonymous

Yes it does, because |x| = 1/3 < 1

12. anonymous

amir that should be =, not <, right?

13. anonymous

like 1/3^n+1-1/1/3-1....then apply limit

14. anonymous

$s _{n} = a/(1-q)$ a is the first term and q is $(1/a ^{n+1})/(1/a ^{n})$

15. anonymous

ya.....amir is right

16. anonymous

thanks! i understand it now!! my teacher didnt explain to us that the sum of the sigma notation equals $x/1-x$

17. anonymous

There's a lot flying around on this post.... for a geometric series, you have that $\sum_{n=0}^\infty = \frac{1}{1-x}$ or, if you start from n = 1, $\sum_{n=1}^\infty \ \frac{x}{1-x}$ In both cases this is valid iff |x| < 1.

18. anonymous

Its the sum of a infinite geometric series.

19. anonymous

was my ans wrong?

20. anonymous

tanusingh I'm not sure exactly what your answer was, could you clarify?

21. anonymous

i actually said that the sum=a^n+1-1/a-1

22. anonymous

Do you mean the nth partial sum?