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

1. anonymous

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

2. anonymous

according to a theorem that all geometric series are convergent.

3. anonymous

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

4. anonymous

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

5. 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.

6. anonymous

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

7. 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?

8. 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).

9. anonymous

*ratio

10. anonymous

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

11. anonymous

1/2

12. anonymous

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

13. anonymous

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

14. anonymous

What's s1 and s2?

15. anonymous

16. anonymous

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

17. 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.

18. anonymous

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

19. 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. =)

20. anonymous

You're welcome :)

21. anonymous

one more, why is $2^{n}/ 2^{n+1} = 1/2$ ?

22. 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$\frac{2^4}{2^5}=\frac{2 \times 2 \times 2 \times 2 }{2 \times 2 \times 2 \times 2 \times 2 }=\frac{1}{2}$

23. anonymous

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

24. anonymous

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

25. anonymous

its always the exponent algebra =/

26. anonymous

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

27. anonymous

def. thanks again =)

28. anonymous

np