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Use the Ratio Test to find the radius of convergence of SUM [x^n/2n^2 ]. So far I have...
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take the limit of an/an-1
\[\lim \frac{x^n}{2n^2}\frac{2(n-1)^2}{x^{n-1}}\]
simplify and then take out all the notn parts
\[\lim x\frac{(n-1)^2}{n^2}\] \[|x|\ \lim \frac{(n-1)^2}{n^2}\to\ |x|*1\] |x|<1 you have a radius of 1
GREAT! I have that!!
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.... if anything, you can always dbl chk with the wolf
haha, "the wolf"!!
Question: does it matter if |x|<1 or \[|x|\le 1\] ??
no, the radius is still 1
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the interval of convergence is -1 to 1
Oooooh
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