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Are you sure about that common ratio?
yep
Okay, so the ratio is greater than 1, which means the series diverges and there is no upper limit to the summation.
so if it is divergent there is no infinite sum?
Correct, a divergent series either increases without bound (approaches \(\infty\)) or oscillates between \(-\infty\) and \(\infty\).
ok can u help with one more? @SithsAndGiggles
sure
The population of a type of local bass can be found using an infinite geometric series where a1 = 72 and the common ratio is \[\frac{ 1 }{ 4 }\]. Find the sum of this infinite series that will be the upper limit of this population.
Alright, this is a convergent series so everything should work nicely. (Convergent because the common ratio is between -1 and 1.) \[\sum_{n=1}^\infty ar^{n-1}\] is the form of the series, and we're given that the first term \(a_1\) is 72. This means \[a_1=\sum_{n=1}^1 ar^{n-1}=ar^{1-1}=72~~\Rightarrow~~a=72\] and so the series is \[\sum_{n=1}^\infty 72\left(\frac{1}{4}\right)^{n-1}\] The sum of this kind of series is given by \[\frac{a}{1-r}\]
is it 96?
Yes
thanks!! :)
yw
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