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anonymous
 one year ago
Consider the power series (n^10(x+7)^n)/((5^n)(n^(32/3)))
need to find the interval of convergence and radius of convergence, please help
anonymous
 one year ago
Consider the power series (n^10(x+7)^n)/((5^n)(n^(32/3))) need to find the interval of convergence and radius of convergence, please help

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freckles
 one year ago
Best ResponseYou've already chosen the best response.0Have you tried applying the ratio test?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have but am not fully understand it yet

freckles
 one year ago
Best ResponseYou've already chosen the best response.0this is what we want: \[\frac{a_{n+1}}{a_n}<1 \] well first we can simplify the left hand expression a bit \[a_{n+1} \cdot \frac{1}{a_n}=\frac{(n+1)^{10} (x+7)^{n+1}}{5^{n+1}( n+1)^\frac{32}{3}} \cdot \frac{5^n n^\frac{32}{3}}{n^{10}}(x+7)^n  \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0for example \[\frac{5^n}{5^{n+1}}=\frac{1}{5} \\ \frac{(x+7)^{n+1}}{(x+7)^n}=x+7\] and yes that (x+7)^n was suppose to be in the bottom

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[a_{n+1} \cdot \frac{1}{a_n}=\frac{(n+1)^{10}(x+7)^{n+1}}{5^{n+1}(n+1)^\frac{32}{3}} \cdot \frac{5^n n^\frac{32}{3}}{n^{10} (x+7)^n} \\ \frac{a_{n+1}}{a_n}=\frac{1}{5}x+7 \cdot \frac{(n+1)^{10} n^\frac{32}{3}}{(n+1)^\frac{32}{3}n^{10}}\] and we have n goes to infinity

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so you need to find the limit of that thing as n goes to infinity and we want that to be less than 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks, can u show me the answer or show me who to get there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to find the interval of convergence and radius of convergence for it

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{5} x+7  \lim_{n \rightarrow \infty } \frac{(n+1)^{10}n^\frac{32}{3}}{(n+1)^{\frac{32}{3}}n^{10}}<1\] well you first need to evaluate that limit I was talking about then you can solve for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so just look at1/5(x+7) how do u eval limit?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0are you saying you don't know how to evaluate this limit? \[\lim_{n \rightarrow \infty} \frac{(n+1)^{10} n^{\frac{32}{3}}}{(n+1)^\frac{32}{3}n^{10}}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \frac{n^{\frac{32}{3}10}}{(n+1)^{\frac{32}{3}10}} \\ =\lim_{n \rightarrow \infty} (\frac{n}{n+1})^\frac{2}{3} \\ = (\lim_{n \rightarrow \infty}\frac{n}{n+1})^\frac{2}{3}\] can you evaluate the limit now ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0yes the inside thing goes to 1 and 1^(2/3) is 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{5} x+7  \lim_{n \rightarrow \infty } \frac{(n+1)^{10}n^\frac{32}{3}}{(n+1)^{\frac{32}{3}}n^{10}}<1 \\ \frac{1}{5} x+7 1<1 \\ \frac{1}{5} x+7<1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0what does that mean ? Solve for x to find the interval of convergence you might want to consider testing the endpoints since at L=1 the ratio test is inconclusive I don't see how you got x=35 when we had an inequality to solve.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0recall f<a where a is positive you have a<f<a

freckles
 one year ago
Best ResponseYou've already chosen the best response.01/5x+7<1 x+7<5 5<x+7<5 can you solve this inequality?

freckles
 one year ago
Best ResponseYou've already chosen the best response.057 is 12 but I don't understand how 57 is 12 also

freckles
 one year ago
Best ResponseYou've already chosen the best response.057 should be 2 since we have 57=(75)=(2)=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay so 2 and 12 for x, and how to find R?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[5<x+7<5 \\ \text{ subtract 7 } \\ 57 <x<57 \\ 12<x<2 \\ \\ \text{ testing the endpoints we will see that the series still diverges so} \\ \text{ the interval of convergence is } 12<x<2 \\ \text{ anyways the diameter is } 2(12)=2+12=10 \\ \text{ the radius is half the diameter }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh got it, thanks so much for your help and patience
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