Question

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`

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`