If #sum_(n=2) ^oo (1+k)^-n=2# what is #k#?
The answer is supposed to be #k=(sqrt3-1)/2# I just don't know how to get there.
The answer is supposed to be
With that,
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The given series is a geometric series with first term ( a = (1 + k)^{-2} ) and common ratio ( r = 1 + k ). For a geometric series to converge, the absolute value of the common ratio must be less than 1. Therefore, ( |1 + k| < 1 ). Solving for ( k ), we have ( -1 < 1 + k < 1 ). Subtracting 1 from each part of the inequality gives ( -2 < k < 0 ). Therefore, ( k ) lies in the interval ( (-2, 0) ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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