geometric series that diverges

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??4/3?^n the sum goes up to infinity and n=0


Answer: geometric series and r=4/3>1 implies series diverges.


Why is the answer 4/3 and not 1 if any number raised to the power of 0 is 1.


Thank you very much for your help.
 
r is the common ratio, and it is what is being raised to a power. To be more thorough, one should say the absolute value of r, |r|, is > than 1, because it applies to the case for negative ratios. i.e. (-4/3)^n. I am not sure what you are talking about in your post, but the reasoning is correct.
 
n=0 not 1 so i thought that (4/3)^n is (4/3)^0 and this equals 1 not 4/3

i want to know why the answer is 4/3 and not 1
 
\(\displaystyle \sum_{n=0}^{\infty}\bigg(\frac{4}{3}\bigg)^n \ and \ \sum_{n=1}^{\infty}\bigg(\frac{4}{3}\bigg)^n \ = \ \infty\)
 
so n=0, means that 4/3 is the value (not raised to any power). In other words, 4/3 is the first term?

If n=1 then I will have to raise 4/3 to the first power, which will also give the same answer?

Is this correct?
 
\(\displaystyle \bigg(\frac{4}{3}\bigg)^0 \ = \ 1, \ not \ \frac{4}{3}.\)

\(\displaystyle \bigg(\frac{4}{3}\bigg)^1 \ = \ \frac{4}{3}.\)
 
Again, I have no idea why you are evaluating the first term if you're just testing for convergence. The fact that it converges has NOTHING to do with n=0, n=1, etc. It is only the ratio that matters. To find the sum you need to evaluate the first term and the sum will be that first term divided by (1-r).
 
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