anonymous
  • anonymous
Find the interval of convergence of the sum of (x-2)^n divided by the square root of n.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
The answer is [1,3), how did they get to that?
anonymous
  • anonymous
I'm doing the problem...
anonymous
  • anonymous
Do you know how to find the biggest chunk of the interval of convergence?

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anonymous
  • anonymous
That is, I'm asking if it's only the end points you're having trouble with.
anonymous
  • anonymous
?
anonymous
  • anonymous
oh sorry, i actually figured out what i did wrong. Thanks tho!
anonymous
  • anonymous
while you're hear though.... how do you do term by term multiplication in a series? i have to find the first four nonzero of the MacLaurin series sinx *cosx
anonymous
  • anonymous
Anyway, the series is convergent for x such that\[\lim_{n \rightarrow \infty}\ \left| \frac{(x-2)^{n+1}/\sqrt{n+1}}{(x-2)^n/\sqrt{n}} \right|<1\] by the ratio test, and then check for convergence at each of the end points, x=1 and x=3. Convergent for 1 by alternating series test, and non-convergent for 3 by integral test, say.
anonymous
  • anonymous
You form the Cauchy product.
anonymous
  • anonymous
does anyone know how i can scan my paper
anonymous
  • anonymous
\[\left( \sum_{}{}a_n \right)\left( \sum_{}{}b_n \right)=\sum_{}{}c_n\]where\[c_n=\sum_{k=0}^{n}a_kb_{n-k}\]
anonymous
  • anonymous
does anyone know how i can scan my paper
anonymous
  • anonymous
You need a scanner to scan your paper, mary :)
anonymous
  • anonymous
tell me steps to scan my paper
anonymous
  • anonymous
\[\sum_{n=0}^{\infty}c_n=\sum_{n=0}^{\infty}\left( \sum_{k=0}^{n}a_kb_{n-k} \right)\]
anonymous
  • anonymous
Well, it depends on your software and machinery. They're not all the same. You have to make sure your scanner is connected to your computer, either through a cable or wireless, and use the appropriate software.

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