anonymous
  • anonymous
Power Series: sigma ((-1)^n x^n)/(n+1) as n-> infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_{0}^{\infty} ((-1)^n x^n)/(n+1)\]
anonymous
  • anonymous
i know that im suppose to replace all n's with n+1 and then use ration test but im not sure what steps to take next
anonymous
  • anonymous
Hi donutboy, how are you? Do you know the series representation for ln(1-x)?

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anonymous
  • anonymous
You want to check the convergence of the series or you want to find its sum?
anonymous
  • anonymous
ratio* hi sklee, im fine, no i dont know the ln(1-x)
anonymous
  • anonymous
im looking for the radius of convergence and interval of convergence
anonymous
  • anonymous
ok great, if you want to find the radius and interval of convergence, then you dont have to know that series representation
anonymous
  • anonymous
For radius of convergence, i assume you will use the Ration test, and not using the formula. So I will show you the work using the test.
anonymous
  • anonymous
Let a_n = (-1)^n x^n / (n+1), then by the Ratio test, lim |a_(n+1)/a_n| = lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)] = lim |x {(n+1)/(n+2)}| Because lim (n+1)/(n+2) = 1 and the series converges if the value of the limit of |a_(n+1)/a_n| is less than 1, we get |x|<1
anonymous
  • anonymous
ahh icic thank you so much! my prof. goes way to fast and its hard to keep up
anonymous
  • anonymous
So the radius of convergence is 1, and the interval we are considering is -1 < x < 1. So we need to check the convergence of the series at the end point x =-1 and x =1
anonymous
  • anonymous
at x = -1, the series become \sum {1/(n+1)}, which is divergent as it is a variation of harmonic series. at x=1, the series becomes \sum {(-1)^n / (n+1)}, which is a variation of the Alternating harmonic series, and this series converges. So the interval of convergence is -1 < x =< 1
anonymous
  • anonymous
quesiton on the lim |a_(n+1)/a_n| = lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)] = lim |x {(n+1)/(n+2)}| part what happens during lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)] ?
anonymous
  • anonymous
x^(n+1)/x^n can be simplified to become x
anonymous
  • anonymous
and i just arrange the term (n+1) and (n+2) as a fraction
anonymous
  • anonymous
oh okay thank you again
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
\[\sum_{0}^{\infty} (-1)^n (x^(2n)/ (2n!)\]
anonymous
  • anonymous
so far ive gotten to \[\lim_{n \rightarrow \infty} \left| (2n!(x^(2n+1)/x^(2n)(2n+2) \right|\]
anonymous
  • anonymous
ok, very good
anonymous
  • anonymous
now the [x^(2n+1)]/x^(2n) is simplified to x
anonymous
  • anonymous
and (2n)! / (2n+2)! is simplified to 1/[(2n+1)(2n+2)] because (2n+2)! = (2n+2)(2n+1)[(2n)!]
anonymous
  • anonymous
alrighty
anonymous
  • anonymous
the limit of 1/[(2n+1)(2n+2)] is 0, so the series converges for all x
anonymous
  • anonymous
so the radius is infinity and the interval of convergence is (-infinity, infinity)
anonymous
  • anonymous
yup, you are right!
anonymous
  • anonymous
thanks! im starting to get this now!
anonymous
  • anonymous
great, you can do it donutboy. All the best!
anonymous
  • anonymous
thanks ^__^ youre awesome!

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