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Sum of geometric series

  • Thread starter MathewsMD
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  • #1
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For the question, shouldn't the sum be a(1/1-r) since we know lrl < 1 then that rn → 0 as n → ∞? I just don't quite understand why they wrote the sum is a(r/1-r). Is there a specific reason they did this? This is just a regular geometric series right? Is there any difference since the sum starts at n = 1 instead of n =0?
 

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  • #2
LCKurtz
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Isn't the first term, which you are calling ##a##, equal to ##r## in that sum?
 
  • #3
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Isn't the first term, which you are calling ##a##, equal to ##r## in that sum?
Isn't ## a = \frac {1}{4}## and ##r = \frac{2}{e}##?
 
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Also, for this question, should the interval of convergence be ## [3,3]##? Why exactly did they say ##(-1,1)##?
 

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Isn't ## a = \frac {1}{4}## and ##r = \frac{2}{e}##?
No. a = 1/(2e)

Added: I'm talking about the first series in the image.
 
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For the question, shouldn't the sum be a(1/1-r)
When you write fractions as text on a single line, you need to be careful about parentheses. 1/1-r means (1/1) - r, and not 1/(1 - r) as you intended.

You have it below, as well.
since we know lrl < 1 then that rn → 0 as n → ∞? I just don't quite understand why they wrote the sum is a(r/1-r). Is there a specific reason they did this? This is just a regular geometric series right? Is there any difference since the sum starts at n = 1 instead of n =0?
 
  • #7
LCKurtz
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Isn't ## a = \frac {1}{4}## and ##r = \frac{2}{e}##?
In the picture where he has written ##\frac 1 4\sum_{k=1}^\infty r^k## the 1/4 is factored out and the remaining sum has ##a=r=\frac 2 e##.
 
  • #8
Ray Vickson
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For the question, shouldn't the sum be a(1/1-r) since we know lrl < 1 then that rn → 0 as n → ∞? I just don't quite understand why they wrote the sum is a(r/1-r). Is there a specific reason they did this? This is just a regular geometric series right? Is there any difference since the sum starts at n = 1 instead of n =0?
Of course there is a difference! If you start the sum at ##n = 0## the first term is ##a##; if you start it at ##n = 1## the first term is ##ar##. The two sums differ by ##a##.

Anyway, they didn't write ##a(r/1-r) = ar - ar = 0##; they wrote ##ar/(1-r)##, where parentheses make all the difference in the world.
 
  • #9
Fredrik
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The sum (##\sum_k r^k##) is 1/(1-r) when the summation starts at 0. If it starts at 1, you can do this: $$\sum_{k=1}^\infty r^k =r\sum_{k=1}^\infty r^{k-1} =r\sum_{k=0}^\infty r^k.$$
 
  • #10
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The sum (##\sum_k r^k##) is 1/(1-r) when the summation starts at 0. If it starts at 1, you can do this: $$\sum_{k=1}^\infty r^k =r\sum_{k=1}^\infty r^{k-1} =r\sum_{k=0}^\infty r^k.$$
So if it is instead ## ∑_{n=3}^∞r^n## then this equals ##∑_{n=0}^∞r^{n+3} = r^3∑_{n=0}^∞r^{n}## so then the sum is ##\frac{r^3}{1-r}## correct?
 
  • #11
LCKurtz
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So if it is instead ## ∑_{n=3}^∞r^n## then this equals ##∑_{n=0}^∞r^{n+3} = r^3∑_{n=0}^∞r^{n}## so then the sum is ##\frac{r^3}{1-r}## correct?
Yes. This agrees with the formula ##\frac a {1-r}## since the first term, ##a##, in ##\sum_{n=0}^∞r^{n+3}## is ##r^3##, as is the first term in ##\sum_{n=3}^∞r^n##. It doesn't matter how you write it.
 

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