difference of two geometric series.
infinity(summation symbol) n=1
[(0.8)^(n-1) - (0.3)^n]

Question

difference of two geometric series.
infinity(summation symbol) n=1
[(0.8)^(n-1) - (0.3)^n]

aravindg · Answer

can u pls se eqn editor?

anonymous · Answer

$$\sum_{n=1}^{\infty}(0.8)^{n-1}-(0.3)^n$$

anonymous · Answer

$$\sum_{n=1}^{\infty} [(0.8)^{n-1} -(0.3)^{n}]$$

aravindg · Answer

i suggest u open the brackets and evaluate sigma separatel and then add

anonymous · Answer

I get (1-0.3)+ (0.8-0.9)
Which is wrong, obviously.So do I find first find what both series converge to then subtract them?

anonymous · Answer

Never mind. I think I got it

anonymous · Answer

Still struggling...I guess the 1-.3 is corrected but I'm not quite sure what to put in the denominator....please help

aravindg · Answer

anonymous · Answer

I have the solution to the question. it's $$1/(1-0.8) -0.3/(1-0.3)  = 5- 3/7 = 32/7   $$
BUT WHY?

aravindg · Answer

wait a sec ..lemme do this on paper then i will post my answer

anonymous · Answer

Thanks

aravindg · Answer

yep !!

aravindg · Answer

:(

aravindg · Answer

I AM TERRIBLY SAD !!

aravindg · Answer

I WROTE A BIG ANSWER AND SUDDENLY GOOGLE CHROME SHOWS "AW SNAP "RELOAD THE PAGE :(

anonymous · Answer

Ohhh I'm so sorry....darn it Google Chrome!!!!

aravindg · Answer

i typed it so nicely for u :( .....anyway i will try to draw the answer i cant bare another latex writing

anonymous · Answer

sure, anything will help. Thanks for the effort

aravindg · Answer

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