Question

find the radius of convergence of the power series ∑ {n=0,∞} [(2^−n) +(3^−n)]z^n

Answer 1

determine the limit of the ration of a_(n+1)/a_n for starters

Answer 2

does that "z" have any special importance? or is it just a generic variable?

Answer 3

the sum is not dependent on z, in essence 'z' is fixed

Answer 4

I can jump straight into using either the nth root test or the ratio test, but the problem is i don't know how to determine if a power series is bounded or not.....please help

Answer 5

\[\lim_{n\to inf}\frac{ \left[2^{-(n+1)} +3^{-(n+1)}\right]z^{n+1}}{\left[2^{-(n)} +3^{-(n)}\right]z^{n}}\]
\[|z|\lim_{n\to inf}\frac{ 2^{-(n+1)} +3^{-(n+1)}}{2^{-(n)} +3^{-(n)}}\]

Answer 6

i would say the Lhop would help to define the limit since this goes to 0/0

Answer 7

\[|z|\lim_{n\to inf}\frac{ 2\cdot2^{-(n)} +3\cdot3^{-(n)}}{2^{-(n)}+3^{-(n)}}\]
\[|z|\lim_{n\to inf}\frac{ 2\cdot2^{-(n)}... +3\cdot3^{-(n)}...}{2^{-(n)}ln(2) +3^{-(n)}ln(3)}\]
then again, maybe not :)

Answer 8

the wolf says the limit of that is 1/2, but i cant really see a way to approach that

if you can determine that step :)

\[|z|\frac12<1\]\[|z|<2~:~Radius=2\]

Answer 9

Yea that is the correct ans, but I dont understand how to approach it myself. Btw the power is to 'n', and not 'n+1'. Thanks very much for you help anyway

Answer 10

the ratio is n+1/n