Find all values of x for which the series converges, and find the sum of the series for those values of x.

Question

anonymous · Answer

1/(x^2)+4/(x^3)+16/(x^4)+64/(x^5)+256/(x^6)

anonymous · Answer

converges for x < ? or x> ?

perl · Answer

what is the original series

perl · Answer

a series has infinite terms, does this series go on

anonymous · Answer

yes :)

perl · Answer

ok it looks like your series is 
$$
\Large{\sum_{n=0}^{\infty} \frac{4^{n}}{x^{n+2}}             }
$$

perl · Answer

to find where it converges , we can use ratio test

perl · Answer

alternatively, you can rewrite the series

perl · Answer

ok it looks like your series is 
$$
\Large{\sum_{n=0}^{\infty} \frac{4^{n}}{x^{n+2}} 
=
   \sum_{n=0}^{\infty} \frac{4^{n}}{x^{n}\cdot x^2}=    \sum_{n=0}^{\infty} (\frac{4}{x})^n\cdot \frac{1}{x^2} \
=        }
$$

anonymous · Answer

i got it ! thank you :)