Answer: (i) \(u_n=4\left(\dfrac54\right)^n+3\).

(ii) The series is convergent for \(-\dfrac45\lt x\lt\dfrac45\).

(iii) \(a=\dfrac{\sqrt5}{2}\) and \(b=2\sqrt5\).

First prove the formula by induction.

For \(n=1\),

\(4\left(\dfrac54\right)^1+3=5+3=8=u_1,\)

so the result is true for \(n=1\).

Assume that for some positive integer \(k\),

\(u_k=4\left(\dfrac54\right)^k+3.\)

Then

\(u_{k+1}=\dfrac{5u_k-3}{4}=\dfrac{5\left(4\left(\dfrac54\right)^k+3\right)-3}{4}.\)

Simplifying,

\(u_{k+1}=\dfrac{20\left(\dfrac54\right)^k+12}{4}=5\left(\dfrac54\right)^k+3=4\left(\dfrac54\right)^{k+1}+3.\)

Therefore the formula holds for all positive integers \(n\).

Now

\(u_r-3=4\left(\dfrac54\right)^r.\)

The infinite series is

\(\sum_{r=1}^{\infty}(u_r-3)x^r=\sum_{r=1}^{\infty}4\left(\dfrac54\right)^r x^r=\sum_{r=1}^{\infty}4\left(\dfrac{5x}{4}\right)^r.\)

This is geometric and converges when

\(\left|\dfrac{5x}{4}\right|\lt 1,\)

so

\(-\dfrac45\lt x\lt\dfrac45.\)

For the logarithmic sum,

\(\ln(u_n-3)=\ln\left(4\left(\dfrac54\right)^n\right)=\ln4+n\ln\dfrac54.\)

Hence

\(\sum_{n=1}^{N}\ln(u_n-3)=N\ln4+\left(\sum_{n=1}^{N}n\right)\ln\dfrac54.\)

Since \(\sum_{n=1}^{N}n=\dfrac{N(N+1)}{2}\),

\(\sum_{n=1}^{N}\ln(u_n-3)=\dfrac{N^2}{2}\ln\dfrac54+N\left(\ln4+\dfrac12\ln\dfrac54\right).\)

So

\(N^2\ln a=\dfrac{N^2}{2}\ln\dfrac54\), giving \(a=\sqrt{\dfrac54}=\dfrac{\sqrt5}{2}\),

and

\(N\ln b=N\ln\left(4\sqrt{\dfrac54}\right)\), giving \(b=4\cdot\dfrac{\sqrt5}{2}=2\sqrt5\).