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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.

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

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?

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

You're welcome :)

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

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