Question

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

Answer 1

By the ratio test, letting\[a_n=\frac{1}{2^n}\]then\[\lim_{n \rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|=\lim_{n \rightarrow \infty}\left| \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} \right|=\lim_{n \rightarrow \infty}\frac{2^n}{2^{n+1}}=\lim_{n \rightarrow \infty}\frac{1}{2}=\frac{1}{2}<1\]

Answer 2

according to a theorem that all geometric series are convergent.

Answer 3

If the magnitude of the ratio of successive terms is less than 1, it is convergent.

Answer 4

ok , why does it not ->0 and diverge by nth term test? and thanks!

Answer 5

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.

Answer 6

And 1/2^n does go to zero as n approaches infinity.

Answer 7

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?

Answer 8

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).

Answer 9

*ratio

Answer 10

ok, lastly, what would the ratio be in that case?

Answer 11

1/2

Answer 12

Divide the (n+1)th term by the nth term to see.

Answer 13

ok so the ratio only has to get you from s1 to s2 and not any further?

Answer 14

What's s1 and s2?

Answer 15

The terms in your sequence?

Answer 16

partial sum 1 ( which would be the index) and partial sum 2 (which should equal index*ratio) ?

Answer 17

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.

Answer 18

The terms you test are the terms in the *sequence* that make up the sum.

Answer 19

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. =)

Answer 20

You're welcome :)

Answer 21

one more, why is \[2^{n}/ 2^{n+1} = 1/2 \] ?

Answer 22

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\[\frac{2^4}{2^5}=\frac{2 \times 2 \times 2 \times 2 }{2 \times 2 \times 2 \times 2 \times 2 }=\frac{1}{2}\]

Answer 23

If you have, in general,\[\frac{a^n}{a^m}\]then it's the case that\[\frac{a^n}{a^m}=a^{n-m}\]

Answer 24

In your case,\[\frac{2^n}{2^{n+1}}=2^{n-(n+1)}=2^{-1}=\frac{1}{2}\]since\[a^{-p}=\frac{1}{a^p}\]

Answer 25

its always the exponent algebra =/

Answer 26

Yeah, you need to come to grips with that before you embark on sequences and series.

Answer 27

def. thanks again =)

Answer 28

np