can someone walk me through this, step by step please
\[\sum_{n=0}^{\infty} \frac{ 1 }{ (\sqrt11)^n }\]
Find the sum of the series, if it converges
do you know how to find the sum of a geometric series ? https://en.wikipedia.org/wiki/Geometric_series#Formula
i would really appreciate the help. I am not good with this question at all. and the square root has me thrown off
I know about \[\frac{ a }{ 1-r }\]
and how to test for convergence with limit
and how it is divergent it r is less than -1 or more than 1
in this case, r= 1/sqrt(11) = 0.30511...
a is 1
\[\frac{ 1 }{ 1-\frac{ 1 }{ \sqrt11 } }\] ?
yes. we can play with that to make it look nicer, but that is the answer.
do i multiply sqrt 11 to top and bottom so I can subtract the two fractions?
\[\frac{ 1 }{ \frac{ 11 }{ 11} -\frac{ \sqrt 11 }{ 11 }}\]
yes, and that is \[ \frac{1}{\frac{11- \sqrt{11}}{11}} = \frac{11}{11- \sqrt{11}}\]
we could also get to \[ \frac{11+\sqrt{11}}{10} \]
this is where i get really thrown off... how do i simplify this now?
oh.... the ten... how did you do that?
\[ \frac{11}{11- \sqrt{11}} \cdot \frac{11+\sqrt{11}}{11+\sqrt{11}}\]
this would then be \[\frac{ 1 }{ 10 }(11+\sqrt 11)\]
but I would not bother too much. They all mean the same value, about 1.43166...
hmmm. I thank you kindly sir
so to answer the question, you know it converges because | r | < 1 and you can use the formula you posted up above to find the sum
the problem is, my professor does not want decimals, he wants the fraction form :(
yes?
i meant yes, no question
people generally would use the last form then... no square roots in the denominator (a hold-over from the days before calculators when dividing by square roots was too hard)
lol, they are still too hard for me (obviously) :) thanks a bunch
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