Find the sum of the infinite geometric series in which a1=8 and r= -1/2
use the formula sum = a ----------- 1-r
is it k? or should i give u the ans
Give me the answer.
sry not 16..........it should be 16/3
put limit n tends infi and u'l see myne formula has nothin wrong......no offence
1/2 to the power n where n is infi will only make it smaller
jose ur goin in wrong direction
cmon answer me (1/2)^infinity is smaller or larger
then 1/2
thats not abt getting it determined but u can say that 1/2^inf must be very small..........so it can be done if u neglect that value
but its very important to note my formula works only when r<1........
the same way inf-1 is inf itself ..........:) don't get frustrated on me....... try to understand
=8 / (3/2) = 8 x2/3 = 16/3 thats it, end of story
thanks elecengineer
the sum of a GP is \[\frac{a(1-r^n)}{1-r}\]
if the absolute value of r is less then 1 , then (1-r^n) ---> 1 , as n-> infinity
so you are multiplying by a factor of 1, and thus why it disappears
r^n gets very small if |r|<1 think of 0.01^1000 , thats 0.01 x 0.01 x 0.01 ... etc 1000 times, it quickly approaches zero therefore 1-r^n appraoches 1
elecengineer u r very good at explaining :)
likewise if you had (-0.01)^999 ( an odd power so its not even ) then you would get -0.01 x -0.01x ... etc
that would give a very small negative answer, so 1-r^n would be just a tiny bit bigger than 1 , but it would still approach 1
All the above is also explained here: http://en.wikipedia.org/wiki/Geometric_series Scroll to "formula" section.
or be very close to 1, so in the limit it is neglected
I stated a wrong equation which lead me to a wrong answer. Sorry for the inconvenience. LOL. I really should start proofreading my works. BTW, thanks for the corrections. Sorry Ishaan. :))
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