anonymous
  • anonymous
Help please?? Find the sum of the infinite series 1/3 +4/9 + 16/27+ 64/81+...if it exists
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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geerky42
  • geerky42
Looks like \[\sum\dfrac{4^n}{3^{n+1}}\]
geerky42
  • geerky42
\[\cdots=\dfrac{1}{3}\sum\dfrac{4^n}{3^n} =\dfrac{1}{3}\sum\left(\dfrac{4}{3}\right)^n\]
geerky42
  • geerky42
Now do you think this series exists?

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anonymous
  • anonymous
yes?
geerky42
  • geerky42
Well, this is geometric series, right?
geerky42
  • geerky42
And we have common ratio greater than 1.
anonymous
  • anonymous
yeah but I need to know what the sum of the series is
geerky42
  • geerky42
For any finite geometric series, formula is \[\sum_{i=1}^{n-1} r^i = \dfrac{1-r^n}{1-r}\]Right?
anonymous
  • anonymous
I think so, but what do put in to get the answer? Can you help walk me through the steps? I'm really confused.
geerky42
  • geerky42
Do you know calculus?
anonymous
  • anonymous
no
geerky42
  • geerky42
Really? ok well just take a look at \[\dfrac{1-r^n}{1-r}\] Since common ratio (r) is greater than 1, what would happen to it if n getting bigger and bigger? \(r^n\) would get bigger and bigger, right?
anonymous
  • anonymous
yes
geerky42
  • geerky42
well, so when you "get" to infinity. You would have infinity in numerator.
geerky42
  • geerky42
So this series doesn't exist.
geerky42
  • geerky42
Does that make sense?
anonymous
  • anonymous
Okay thank you <3

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