Tweaking the Fibonacci Sequence!

It is clear that $B_n=k^{n+1}F_n$ (check by direct substitution!), where $F_n$ is the $n$ th Fibonacci number. Now it is well-known that the Generating function of the Fibonacci series is given by : $\sum_{n=0}^\infty F_nz^n=\frac{1}{1-z-z^2}$ within the radius of convergence $|z|<\frac{1}{\phi}$ . Hence for all $|kz|< \frac{1}{\phi}$ , we have : $\sum_{n=0}^\infty B_nz^n=k\sum_{n=0}^\infty F_n(kz)^n=\frac{k}{1-kz-k^2z^2}$ Take $k < \frac{1}{\phi}$ and hence the above series is uniformly convergent in the interval $(0,1)$ . Thus we can integrate both sides of the power series to yield $\sum_{n=0}^\infty \frac{B_n}{n+1}= \int_{0}^{k} \frac{1}{1-x-x^2} dx=\frac{1}{\sqrt{5}}\log \bigg(\frac{1+k(\sqrt{5}-1)/2}{1-k(\sqrt{5}+1)/2}\bigg)$ which yields the result.

Note : I doubt whether the series is convergent for $k>\frac{1}{\phi}$ . Hence the statement is the problem needs to be modified by restricting $0<k<\frac{1}{\phi}$ , where $\phi$ is the Golden ratio.