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Simple Series

  1. Jan 17, 2008 #1

    For example 26, I don't see how (for the 1st part) n=9. In the definition of the summing of them it says n is the number above SIGMA, so why not 10? Is it to do with n=1, which isn't k=0 from the definition?
  2. jcsd
  3. Jan 17, 2008 #2


    I did the 1st, which diverged by the divergence test.

    But the 2nd, I tried the ratio test, but it got far 2 complicated too early on. Which test do I use? I tried them all, what am I doing wrong?
  4. Jan 17, 2008 #3


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    Yes, the formula given for a geometric series starts with k= 0. The example given
    [tex]\sum_{k= 1}^10 \frac{2}{3^n}[/tex]
    starts with k= 1. The simplest way to handle that is to factor 1/3 out of the product:
    [tex]\sum_{k=1}^{10}\frac{2}{3} \frac{1}{3^{k-1}}[/tex]
    Now, let j= k-1. Then k= 1 becomes j= 1-1= 0 and k= 10 becomes j= [itex]j= 10-1= 9[/itex]. Of course, [itex]3^{k-1}[/itex] becomes [itex]3^j[/itex]. The sum is now
    [tex]\sum_{j= 0}^9 \frac{2}{3}\frac{1}{3^j}[/itex]
    That is a geometric series with A= 2/3, r= 1/3, and n= 9.
  5. Jan 17, 2008 #4


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    For that one,
    [tex]\lim_{x\rightarrow 0}\frac{e^x- x- 1}{sin^2(x)}[/itex]
    since both numerator and denominator go to 0 I would use L'Hopital's rule (twice). And, again, you don't need to find the limit itself!
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