Convergent or divergent 16-8+4-...
do you know what kinda series that is?
hint: is there a common ratio?
if so what is it?
No, I have to figure out the infinite geometric series
there is a common ration since it is a geometric series
the geometric series here has a ratio of (-1/2)
do you see that now?
has a common ratio of (-1/2)*
Yes
So is it divergent?
now since you just need to determine if it is divergent or convergent you just need to tell me if you have |r|<1 --- if |r|<1 then it is convergent if |r|>=1 then it is divergent
you have r=-1/2 so which inequality is true ?
oh, its convergent. <
yep yep
thank you! can you answer another?
okay
Find the sum of each infinite series, if it exists 440+220+110+...
let's see what kinda series is this?
can you tell?
have you looked for a common ratio?
i think its an infinite one
if there is a common ratio, it is a geometric series
i looked for one but really couldnt figure it out
220/440=? 110/220=?
okay so 1/2
yes both of them are equal to 1/2 therefore there is a common ratio of 1/2 therefore since it has a common ratio it is a geometric series
use the same test as we did before
you mean the convergent and divergent thing?
if so its convergent since its less than 1
the geometric convergeny test: if |r|<1 then it is convergent if |r|>=1 then it is divergent
and yes
if we had 2,4,8,... this would be a different story
the common ratio here would be 2
so 2+4+8+... would diverge
and divergent since its over 1?
right
okay yeah i get it.! so if i were to find the sum of the last series? how would i do that?
if we have -2,4,-8,... then the common ratio is -2 and so -2+4+(-8)+... is divergent since |-2|=2>1
the formula is this: \[\sum_{n=0}^{\infty}a r^n=\frac{a}{1-r}\]
so you have a=440 and r=1/2
enter those in
hmmm cant seem to find it on my calculator. D:
find what/
the formula to plug the numbers into
\[\frac{440}{1-\frac{1}{2}} =?\] a division sign and subtraction sign?
(440) div (1-1 div 2)
so you think your calculator has the formula in it?
what kinda calculator do you have?
TI 84 Plus by Texas Instruments
it looks like you can use the calculator to by pass using the formula
but you have to know the put it in sigma notation
440,220,110,... first number is 440 440,440(1/2),440(1/2)^2,... this can be written as \[\sum_{k=0}^{\infty}440 \cdot (\frac{1}{2})^k\] so it says you can enter seq(440(1/2)^k, 0,inf) it looks like
then you can use the sum command
sum(Ans)
that seems like a lot of work
I think using the a/(1-r) is pretty easy
okay yeah that way seemed a bit easier
so does that big greek looking E have any significance?
is that what you are confused about?
that big sigma thing means we are adding a whole bunch of terms
example: \[\sum_{k=0}^{5}k^2 \text{ means } 0^2+1^2+2^2+3^2+4^2+5^2\]
the sigma sign tells us we are adding the k=0 underneath means we going to starting adding terms starting with k=0 the 5 on top means that is where we will stop
well i missed class today. Then learned a new section and when i got my work after school she gave me a 40 problem packet and told me it was due tomorrow. so im just now trying to learn what all these words and symbols mean.
so instead of writting the series in sigma notation I could have wrote: \[a+ar+ar^2+ar^3+ar^4 + \cdots =\frac{a}{1-r}\]
where |r|<1
so to do these it has to be convergent.?
yes you can only use that formula if you have |r|<1
ok sounds good
ok so when i plug in the numbers i got 880. is that the final answer?
well if youre still there. you were a big help thank you and sorry im a dumbass
880 is correct
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