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.

Answer 1

#k=(sqrt(3)-1)/2#

A geometric series of the form #sum_(n=0)^oor^n# with #|r|<1# evaluates to
#sum_(n=0)^oor^n = 1/(1-r)#

With that,

#sum_(n=2)^oo(1+k)^(-n) = sum_(n=2)^oo(1/(1+k))^n#
#=-(1/(1+k))^0-(1/(1+k))^1+sum_(n=0)^oo(1/(1+k))^n#
#=-1-1/(1+k)+1/(1-(1/(1+k)))#
#=-1-1/(1+k)+1/(k/(1+k))#
#=-1-1/(1+k)+(1+k)/k#
#=2#
We can now solve for #k#. Multiplying through by #k(1+k)#, we get
#-k(1+k) - k + (1+k)^2 = 2k(1+k)#
#=> -k-k^2-k+1+2k+k^2 = 2k^2+2k#
#=> 1 = 2k^2+2k#
#=> 2k^2+2k-1 = 0#
#=> k = (-1+-sqrt(3))/2#
But we must have #k>0 or k<-2# for #lim_(n->oo)(1+k)^(-n)=0#, a necessary condition for convergence, thus our only possibility becomes the positive option:
#k=(sqrt(3)-1)/2#
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Answer 2

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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Answer from HIX Tutor

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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