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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} 1/3^n\] please help!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is the sum of a geometric sequence: \[\sum_{n=1}^\infty x^n = \frac{x}{1x} \] so long as \[x < 1\] What is x in this case?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that'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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left(\frac{1}{3}\right)^n = \frac{1}{3^n} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is the case of convergence?jemurray

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does that mean that there is no x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It means x=1/3. What about convergence, tanusingh?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like does this infinte series converge as n tends to infinitey

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty}1/3^{n} = (1/3)/(1(1/3))<1/2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u can also use the formula a^n+11/a1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes it does, because x = 1/3 < 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0amir that should be =, not <, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like 1/3^n+11/1/31....then apply limit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[s _{n} = a/(1q)\] a is the first term and q is \[(1/a ^{n+1})/(1/a ^{n})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks! i understand it now!! my teacher didnt explain to us that the sum of the sigma notation equals \[x/1x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There's a lot flying around on this post.... for a geometric series, you have that \[\sum_{n=0}^\infty = \frac{1}{1x} \] or, if you start from n = 1, \[\sum_{n=1}^\infty \ \frac{x}{1x} \] In both cases this is valid iff x < 1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its the sum of a infinite geometric series.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tanusingh I'm not sure exactly what your answer was, could you clarify?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i actually said that the sum=a^n+11/a1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you mean the nth partial sum?
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