anonymous
  • anonymous
why does the series {1/2^n} from n=1 to infinity converge?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
By the ratio test, letting\[a_n=\frac{1}{2^n}\]then\[\lim_{n \rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|=\lim_{n \rightarrow \infty}\left| \frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}} \right|=\lim_{n \rightarrow \infty}\frac{2^n}{2^{n+1}}=\lim_{n \rightarrow \infty}\frac{1}{2}=\frac{1}{2}<1\]
anonymous
  • anonymous
according to a theorem that all geometric series are convergent.
anonymous
  • anonymous
If the magnitude of the ratio of successive terms is less than 1, it is convergent.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.