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

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

it is the case of convergence?jemurray

does that mean that there is no x?

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

like does this infinte series converge as n tends to infinitey

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

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

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

amir that should be =, not <, right?

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

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

ya.....amir is right

Its the sum of a infinite geometric series.

was my ans wrong?

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

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

Do you mean the nth partial sum?