Math2400
  • Math2400
determine the sum of this series...i thought it was divergent but I guess it diverges cuz it says that it's not divergent
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Math2400
  • Math2400
\[\sum_{n=1}^{\infty} \frac{ (-3)^{n-1} }{ 5^n}\]
Math2400
  • Math2400
@jim_thompson5910 would u be able to help me on this please?
ganeshie8
  • ganeshie8
As a start, write the fraction in single exponent

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ganeshie8
  • ganeshie8
\[\sum_{n=1}^{\infty} \frac{ (-3)^{n-1} }{ 5^n} = \sum_{n=1}^{\infty} \frac{ (-3)^{n} }{ (-3)5^n} = -\frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{ -3 }{ 5}\right)^n\]
ganeshie8
  • ganeshie8
next recall geometric series
Math2400
  • Math2400
a/(1 - r) ??
ganeshie8
  • ganeshie8
Yep
Math2400
  • Math2400
what would be r? this is where I'm confused
ganeshie8
  • ganeshie8
r = the stuff under the exponent = -3/5
Math2400
  • Math2400
and for a i just plug in 1 into the equation?
Math2400
  • Math2400
and get that value?
ganeshie8
  • ganeshie8
there is no equation
ganeshie8
  • ganeshie8
for "a", just plugin n=1 into the general term of series
Math2400
  • Math2400
that's what i mean haha :)
ganeshie8
  • ganeshie8
i kno, just making sure you see the difference between "equation" and "expression"
Math2400
  • Math2400
okay i plugged everything in but it's zero??
ganeshie8
  • ganeshie8
what do you get for first term, a ?
Math2400
  • Math2400
(-3)^(1-1)/5^1=0
ganeshie8
  • ganeshie8
Ah no, below is our slightly massaged series : \[-\frac{1}{3}\color{blue}{\sum_{n=1}^{\infty} \left(\frac{ -3 }{ 5}\right)^n}\] plugin \(n=1\), you get \(a_1 = -\frac{3}{5} \) \(r = -\frac{3}{5}\) so the infinite sum is \[\dfrac{-3/5}{1-(-3/5)}\] simplify
Math2400
  • Math2400
-0.375
Math2400
  • Math2400
that's still not right tho haha
ganeshie8
  • ganeshie8
looks good, don't forget that -1/3 in front
Math2400
  • Math2400
got it! thanks :)
ganeshie8
  • ganeshie8
np

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