Question

please help!! How do you solve this sigma notation with an infinite series?

Answer 1

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

Answer 2

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?

Answer 3

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.

Answer 4

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

Answer 5

it is the case of convergence?jemurray

Answer 6

does that mean that there is no x?

Answer 7

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

Answer 8

like does this infinte series converge as n tends to infinitey

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

amir that should be =, not <, right?

Answer 13

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

Answer 14

\[s _{n} = a/(1-q)\]

a is the first term
and q is \[(1/a ^{n+1})/(1/a ^{n})\]

Answer 15

ya.....amir is right

Answer 16

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

Answer 17

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.

Answer 18

Its the sum of a infinite geometric series.

Answer 19

was my ans wrong?

Answer 20

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

Answer 21

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

Answer 22

Do you mean the nth partial sum?