anonymous
  • anonymous
could you also help me with (-1)^(n+1)(11^(n-1))/((n+7)^6(10^(n+2)) for L and where the equation editor is?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Loser66
  • Loser66
I don't get what you mean by where the equation editor is? Your problem : \[L=\dfrac{(-1)^{n+1}*(11^{n-1})}{(n+7)^6*(10^{n+2})}\] right? and what are we supposed to do?
anonymous
  • anonymous
find L value
anonymous
  • anonymous
thanks!

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Loser66
  • Loser66
find L value for n =??
anonymous
  • anonymous
yes
anonymous
  • anonymous
looking to find the value of L in the ratio test
Loser66
  • Loser66
You meant test whether the sequence converges or diverges?
anonymous
  • anonymous
yes
jtvatsim
  • jtvatsim
Not necessarily related to your problem, but you asked where the equation editor is, you'll find it here: (see attached)
Loser66
  • Loser66
@jtvatsim please help him, I forgot how to do ratio test. :)
jtvatsim
  • jtvatsim
Ah... I was hoping you knew... Let me dust off the Calculus part of my brain. If I recall, it's pretty straightforward. :)
jtvatsim
  • jtvatsim
The ratio test says that we evaluate \[L = \lim_{n \rightarrow \infty} | \frac{a_{n+1}}{a_n} |\]
jtvatsim
  • jtvatsim
The criteria are that if L < 1 the series converges absolutely, if L > 1 the series diverges, and if L = 1, we have no clue.
jtvatsim
  • jtvatsim
So, with your current question, we have a very ugly fraction to take care of, but let's see what we get...
jtvatsim
  • jtvatsim
I'm not going to type out the gigantic intermediate steps, but I will clarify if you get an answer different than mine.
jtvatsim
  • jtvatsim
OK, I've gotten to this point so far
jtvatsim
  • jtvatsim
\[\lim_{n \rightarrow \infty} | - \frac{11 (n+7)^6}{10 (n+8)^6} |\]
jtvatsim
  • jtvatsim
Is that somewhat similar to what you got after simplifying?
anonymous
  • anonymous
yes you don't have to do the steps just struggling with may answer part, i have gotten many problems right so far but for some reason can't get this one
anonymous
  • anonymous
any updates???

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