If sum_(n=2) ^oo (1+k)^-n=2∞∑n=2(1+k)−n=2 what is kk?

The answer is supposed to be k=(sqrt3-1)/2k=√3−12 I just don't know how to get there.

1 Answer
sente
Dec 14, 2016

k=(sqrt(3)-1)/2k=√3−12

Explanation:

A geometric series of the form sum_(n=0)^oor^n∞∑n=0rn with |r|<1|r|<1 evaluates to

sum_(n=0)^oor^n = 1/(1-r)∞∑n=0rn=11−r

With that,

sum_(n=2)^oo(1+k)^(-n) = sum_(n=2)^oo(1/(1+k))^n∞∑n=2(1+k)−n=∞∑n=2(11+k)n

=-(1/(1+k))^0-(1/(1+k))^1+sum_(n=0)^oo(1/(1+k))^n=−(11+k)0−(11+k)1+∞∑n=0(11+k)n

=-1-1/(1+k)+1/(1-(1/(1+k)))=−1−11+k+11−(11+k)

=-1-1/(1+k)+1/(k/(1+k))=−1−11+k+1k1+k

=-1-1/(1+k)+(1+k)/k=−1−11+k+1+kk

=2=2

We can now solve for kk. Multiplying through by k(1+k)k(1+k), we get

-k(1+k) - k + (1+k)^2 = 2k(1+k)−k(1+k)−k+(1+k)2=2k(1+k)

=> -k-k^2-k+1+2k+k^2 = 2k^2+2k⇒−k−k2−k+1+2k+k2=2k2+2k

=> 1 = 2k^2+2k⇒1=2k2+2k

=> 2k^2+2k-1 = 0⇒2k2+2k−1=0

=> k = (-1+-sqrt(3))/2⇒k=−1±√32

But we must have k>0 or k<-2k>0ork<−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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