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.

1 Answer
Dec 14, 2016

Answer:

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

Explanation:

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#