Finding sum of infinite geometric series

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SUMMARY

The infinite geometric series presented is 1 - √2 + 2 - 2√2 + ..., where the common ratio (r) is determined to be 2. According to the formula for the sum of an infinite geometric series, S = 1 / (1 - r), the series diverges since |r| = 2 is greater than 1. Therefore, the correct conclusion is that the series diverges, confirming option c as the answer. The calculations leading to this conclusion were verified by multiple participants in the discussion.

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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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