anonymous
  • anonymous
The interval of convergence of the series [(3^n(x-2)^n)/ (n!)] is from n=0 to infinity.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[ \sum_{n=0}^{\infty} [(3^n(x-2)^n/ n!)]\]
anonymous
  • anonymous
To be clear numerator: 3^n(x-2)^n Denominator: n!
amistre64
  • amistre64
nice :)

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anonymous
  • anonymous
im so bad with series questions...i dont even know where to start with some of them
amistre64
  • amistre64
i recall this vaguely, so ill have to resort to pauls workings of it
anonymous
  • anonymous
ok
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx if you wanna make sure i read it right ;)
amistre64
  • amistre64
it looks like we take the ratio like in a ratio test
anonymous
  • anonymous
yeah i rmbr my teaching saying if the series has a factorial and a power u use the ratio test
amistre64
  • amistre64
\[\sum_{n=0}^{\infty} \frac{3^n(x-2)^n}{n!}\] \[\lim_{n\to inf} \frac{3^n(x-2)^n}{n!} \frac{(n-1)!}{3^{(n-1)}(x-2)^{(n-1)}}\] \[\lim_{n\to inf} \frac{3(x-2)}{n}\]
amistre64
  • amistre64
as n get large; the value goes to zero
anonymous
  • anonymous
i have a qustion
amistre64
  • amistre64
which I believe means that it is convergent everywhere
anonymous
  • anonymous
i thought for the ratio test ur supppsed to add 1 to n not subtract
anonymous
  • anonymous
so like it shud be 3^(n+1)* (x-2)^n+1
amistre64
  • amistre64
the ration test is of the form\[\frac{A_{now}}{A_{past}}\]or\[\frac{A_{future}}{A_{now}}\]
amistre64
  • amistre64
the idea being that your ratio is a proportion of 2 terms with the top being the one right after the bottom
amistre64
  • amistre64
so either n+1 the left part; or n-1 the right part
anonymous
  • anonymous
wat do u mean left and right part
anonymous
  • anonymous
srry my teacher never talked about ratio test like that
amistre64
  • amistre64
\[\lim_{n\to inf} \frac{3^{(n+1)}(x-2)^{(n+1)}}{(n+1)!} \frac{n!}{3^{n}(x-2)^{n}}\] \[\lim_{n\to inf} \frac{3(x-2)}{n+1}\] same results over all
amistre64
  • amistre64
since our ratio is a fraction over a fraction ..... \[\cfrac{\frac{a}{b}}{\frac{c}{d}}\to\cfrac{\frac{a}{b}}{\frac{c}{d}}*\cfrac{\frac{d}{c}}{\frac{d}{c}}= \frac{a}{b}\frac{d}{c}\]
amistre64
  • amistre64
we end up with a left and a right part to this
anonymous
  • anonymous
so now how do we find the interval of converegence since that is wat the question is asking
amistre64
  • amistre64
well, we view the results of our limit ratio
amistre64
  • amistre64
its zero; which tells us that the interval on which this converges is from -inf to inf
amistre64
  • amistre64
example 4 of the link i posted gives a reasoning as to why i believe :)
anonymous
  • anonymous
oh i see. thanks for helping. =)
amistre64
  • amistre64
youre welcome

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