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

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

Mathematics
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By the ratio test, letting\[a_n=\frac{1}{2^n}\]then\[\lim_{n \rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|=\lim_{n \rightarrow \infty}\left| \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} \right|=\lim_{n \rightarrow \infty}\frac{2^n}{2^{n+1}}=\lim_{n \rightarrow \infty}\frac{1}{2}=\frac{1}{2}<1\]
according to a theorem that all geometric series are convergent.
If the magnitude of the ratio of successive terms is less than 1, it is convergent.

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ok , why does it not ->0 and diverge by nth term test? and thanks!
The 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.
And 1/2^n does go to zero as n approaches infinity.
awesome, and i could use the a/1-r with a=1/2 and r=1/2^2 to find the same thing as the ratio test?
If 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).
*ratio
ok, lastly, what would the ratio be in that case?
1/2
Divide the (n+1)th term by the nth term to see.
ok so the ratio only has to get you from s1 to s2 and not any further?
What's s1 and s2?
The terms in your sequence?
partial sum 1 ( which would be the index) and partial sum 2 (which should equal index*ratio) ?
The 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.
The terms you test are the terms in the *sequence* that make up the sum.
ok ok. thanks so much, cleared up a lot of questions i was wasting a lot of time trying to figure out on my own. =)
You're welcome :)
one more, why is \[2^{n}/ 2^{n+1} = 1/2 \] ?
Consider \[\frac{2^4}{2^5}\]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\[\frac{2^4}{2^5}=\frac{2 \times 2 \times 2 \times 2 }{2 \times 2 \times 2 \times 2 \times 2 }=\frac{1}{2}\]
If you have, in general,\[\frac{a^n}{a^m}\]then it's the case that\[\frac{a^n}{a^m}=a^{n-m}\]
In your case,\[\frac{2^n}{2^{n+1}}=2^{n-(n+1)}=2^{-1}=\frac{1}{2}\]since\[a^{-p}=\frac{1}{a^p}\]
its always the exponent algebra =/
Yeah, you need to come to grips with that before you embark on sequences and series.
def. thanks again =)
np

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