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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

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  1. freckles
    • one year ago
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    Have you tried applying the ratio test?

  2. anonymous
    • one year ago
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    i have but am not fully understand it yet

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

  4. freckles
    • one year ago
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    for 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

  5. freckles
    • one year ago
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    \[|a_{n+1} \cdot \frac{1}{a_n}|=|\frac{(n+1)^{10}(x+7)^{n+1}}{5^{n+1}(n+1)^\frac{32}{3}} \cdot \frac{5^n n^\frac{32}{3}}{n^{10} (x+7)^n}| \\ |\frac{a_{n+1}}{a_n}|=\frac{1}{5}|x+7| \cdot |\frac{(n+1)^{10} n^\frac{32}{3}}{(n+1)^\frac{32}{3}n^{10}}|\] and we have n goes to infinity

  6. freckles
    • one year ago
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    so you need to find the limit of that thing as n goes to infinity and we want that to be less than 1

  7. anonymous
    • one year ago
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    okay thanks, can u show me the answer or show me who to get there?

  8. anonymous
    • one year ago
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    to find the interval of convergence and radius of convergence for it

  9. freckles
    • one year ago
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    \[\frac{1}{5} |x+7| | \lim_{n \rightarrow \infty } \frac{(n+1)^{10}n^\frac{32}{3}}{(n+1)^{\frac{32}{3}}n^{10}}|<1\] well you first need to evaluate that limit I was talking about then you can solve for x

  10. anonymous
    • one year ago
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    so just look at1/5(x+7) how do u eval limit?

  11. freckles
    • one year ago
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    are you saying you don't know how to evaluate this limit? \[\lim_{n \rightarrow \infty} \frac{(n+1)^{10} n^{\frac{32}{3}}}{(n+1)^\frac{32}{3}n^{10}}\]

  12. freckles
    • one year ago
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    \[\lim_{n \rightarrow \infty} \frac{n^{\frac{32}{3}-10}}{(n+1)^{\frac{32}{3}-10}} \\ =\lim_{n \rightarrow \infty} (\frac{n}{n+1})^\frac{2}{3} \\ = (\lim_{n \rightarrow \infty}\frac{n}{n+1})^\frac{2}{3}\] can you evaluate the limit now ?

  13. anonymous
    • one year ago
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    is it 1?

  14. freckles
    • one year ago
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    yes the inside thing goes to 1 and 1^(2/3) is 1

  15. freckles
    • one year ago
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    \[\frac{1}{5} |x+7| | \lim_{n \rightarrow \infty } \frac{(n+1)^{10}n^\frac{32}{3}}{(n+1)^{\frac{32}{3}}n^{10}}|<1 \\ \frac{1}{5} |x+7| |1|<1 \\ \frac{1}{5} |x+7|<1\]

  16. anonymous
    • one year ago
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    is this for R or X?

  17. anonymous
    • one year ago
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    is x then 35?

  18. freckles
    • one year ago
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    what 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.

  19. freckles
    • one year ago
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    recall |f|<a where a is positive you have -a<f<a

  20. freckles
    • one year ago
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    1/5|x+7|<1 |x+7|<5 -5<x+7<5 can you solve this inequality?

  21. anonymous
    • one year ago
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    -12 for both?

  22. freckles
    • one year ago
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    -5-7 is -12 but I don't understand how 5-7 is -12 also

  23. freckles
    • one year ago
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    5-7 should be -2 since we have 5-7=-(7-5)=-(2)=-2

  24. anonymous
    • one year ago
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    oh okay so -2 and -12 for x, and how to find R?

  25. freckles
    • one year ago
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    \[-5<x+7<5 \\ \text{ subtract 7 } \\ -5-7 <x<5-7 \\ -12<x<-2 \\ \\ \text{ testing the endpoints we will see that the series still diverges so} \\ \text{ the interval of convergence is } -12<x<-2 \\ \text{ anyways the diameter is } -2-(-12)=-2+12=10 \\ \text{ the radius is half the diameter }\]

  26. anonymous
    • one year ago
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    oh got it, thanks so much for your help and patience

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