anonymous
  • anonymous
please help!! How do you solve this sigma notation with an infinite series?
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\sum_{n=1}^{\infty} 1/3^n\] ---please help!!
anonymous
  • 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?
anonymous
  • 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.

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More answers

anonymous
  • anonymous
\[\left(\frac{1}{3}\right)^n = \frac{1}{3^n} \]
anonymous
  • anonymous
it is the case of convergence?jemurray
anonymous
  • anonymous
does that mean that there is no x?
anonymous
  • anonymous
It means x=1/3. What about convergence, tanusingh?
anonymous
  • anonymous
like does this infinte series converge as n tends to infinitey
anonymous
  • anonymous
\[\sum_{n=1}^{\infty}1/3^{n} = (1/3)/(1-(1/3))<1/2\]
anonymous
  • anonymous
u can also use the formula a^n+1-1/a-1
anonymous
  • anonymous
Yes it does, because |x| = 1/3 < 1
anonymous
  • anonymous
amir that should be =, not <, right?
anonymous
  • anonymous
like 1/3^n+1-1/1/3-1....then apply limit
anonymous
  • anonymous
\[s _{n} = a/(1-q)\] a is the first term and q is \[(1/a ^{n+1})/(1/a ^{n})\]
anonymous
  • anonymous
ya.....amir is right
anonymous
  • anonymous
thanks! i understand it now!! my teacher didnt explain to us that the sum of the sigma notation equals \[x/1-x\]
anonymous
  • 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.
anonymous
  • anonymous
Its the sum of a infinite geometric series.
anonymous
  • anonymous
was my ans wrong?
anonymous
  • anonymous
tanusingh I'm not sure exactly what your answer was, could you clarify?
anonymous
  • anonymous
i actually said that the sum=a^n+1-1/a-1
anonymous
  • anonymous
Do you mean the nth partial sum?

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