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anonymous
 5 years ago
why does the series {1/2^n} from n=1 to infinity converge?
anonymous
 5 years ago
why does the series {1/2^n} from n=1 to infinity converge?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the ratio test, letting\[a_n=\frac{1}{2^n}\]then\[\lim_{n \rightarrow \infty}\left \frac{a_{n+1}}{a_n} \right=\lim_{n \rightarrow \infty}\left \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} \right=\lim_{n \rightarrow \infty}\frac{2^n}{2^{n+1}}=\lim_{n \rightarrow \infty}\frac{1}{2}=\frac{1}{2}<1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0according to a theorem that all geometric series are convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the magnitude of the ratio of successive terms is less than 1, it is convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , why does it not >0 and diverge by nth term test? and thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The nth term test says, if in the limit, your terms DON'T go to zero, then the series WON'T converge. That's all it says.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And 1/2^n does go to zero as n approaches infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome, and i could use the a/1r with a=1/2 and r=1/2^2 to find the same thing as the ratio test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If it's a geometric series with common ration less than 1, you can use the formula, though you have to show first that you do in fact have a geometric series (even though it's obvious here).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, lastly, what would the ratio be in that case?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Divide the (n+1)th term by the nth term to see.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so the ratio only has to get you from s1 to s2 and not any further?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The terms in your sequence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0partial sum 1 ( which would be the index) and partial sum 2 (which should equal index*ratio) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The partial sum for a geometric series is something different to the limiting sum that you use (i.e. your formula). The ratio of two successive partial sums will not, in general, be the common ratio.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The terms you test are the terms in the *sequence* that make up the sum.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok ok. thanks so much, cleared up a lot of questions i was wasting a lot of time trying to figure out on my own. =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one more, why is \[2^{n}/ 2^{n+1} = 1/2 \] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Consider \[\frac{2^4}{2^5}\]The definition of that notation is such that\[2^4=2 \times 2 \times 2 \times 2 \]and\[2^5=2 \times 2 \times 2 \times 2 \times 2 \]so\[\frac{2^4}{2^5}=\frac{2 \times 2 \times 2 \times 2 }{2 \times 2 \times 2 \times 2 \times 2 }=\frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you have, in general,\[\frac{a^n}{a^m}\]then it's the case that\[\frac{a^n}{a^m}=a^{nm}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In your case,\[\frac{2^n}{2^{n+1}}=2^{n(n+1)}=2^{1}=\frac{1}{2}\]since\[a^{p}=\frac{1}{a^p}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its always the exponent algebra =/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, you need to come to grips with that before you embark on sequences and series.
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