Finding sum of infinite geometric series

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Discussion Overview

The discussion revolves around finding the sum of an infinite geometric series defined by the terms 1 - √2 + 2 - 2√2 + ... Participants explore the convergence of the series and the implications of the common ratio.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes that the sum of the series is A (0.414) based on their calculations.
  • Another participant suggests that for the series to be geometric, the common ratio must be identified, indicating that it appears to be 2.
  • Several participants reference the formula for the sum of a geometric series, noting that it converges only if the absolute value of the common ratio is less than 1.
  • There is a repeated assertion that since the common ratio is 2, the series should diverge, leading to the conclusion that the answer is C (the series diverges).

Areas of Agreement / Disagreement

Participants express disagreement regarding the convergence of the series, with some supporting the idea that it diverges while others initially propose a finite sum. The discussion remains unresolved as participants explore different interpretations of the series.

Contextual Notes

Participants have not fully clarified the assumptions regarding the common ratio and its implications for convergence, leading to uncertainty in the conclusions drawn.

fluffertoes
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find the sum of this infinite geometric series:
1 - √2 + 2 - 2√2 + ...

a.) .414
b.) -2.414
c.) series diverges
d.) 2

I found that the common difference is 2, so I calculated this:

S= -.414/-1
s= .414

So i got that the answer is A, but will you check this?
 
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In order for this to be a geometric series, we must have:

$$S=(1-\sqrt{2})\sum_{k=0}^{\infty}\left(a^k\right)$$

Going by the terms given, it appears that $a=2$, and so what does this tell us about the convergence?
 
MarkFL said:
In order for this to be a geometric series, we must have:

$$S=(1-\sqrt{2})\sum_{k=0}^{\infty}\left(a^k\right)$$

Going by the terms given, it appears that $a=2$, and so what does this tell us about the convergence?

oh gosh i don't know
 
fluffertoes said:
oh gosh i don't know

Recall that:

$$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}$$ but only if $|r|<1$

Otherwise, the sum diverges. :D
 
MarkFL said:
Recall that:

$$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}$$ but only if $|r|<1$

Otherwise, the sum diverges. :D

But r stood to equal 2, so the answer should be c?
 
fluffertoes said:
but r stood to equal 2, so the answer should be c?

hello??
 
fluffertoes said:
But r stood to equal 2, so the answer should be c?

Yes.

fluffertoes said:
hello??

Sorry, but I was working on a coding request at vBorg. (Sweating)
 

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