Question

We have string `f_n` with `f_(n+1)-f_n=3f_n`, and `f_n=-2^(2n-1)`. If `h_n=f_1+f_2+...+f_n` then `lim_(n->oo)h_n/h_(n+1)=`?

Answer 1

See below.

Explanation:

If `h_n = -1/2sum_(k=1)^n 2^(2k) = -1/2sum_(k=1)^n 4^k` but

`sum_(k=1)^n 4^k=(4^(n+1)-1)/(4-1)-1 = 4/3(4^n-1)` then

`h_n/h_(n+1) = (4^n-1)/(4^(n+1)-1)` and

`lim_(n->oo)h_n/h_(n+1)=lim_(n->oo)(1-4^(-n))/(4-4^(-n))=1/4`