jmc

Let $$S=\sum _{n=1}^{\infty }\frac{5n-1}{k^n}=\frac{4}{k}+\frac{9}{k^2}+\frac{14}{k^3}+\cdots .$$Multiplying by $k$ gives us $$kS=4+\frac{9}{k}+\frac{14}{k^2}+\frac{19}{k^3}+\cdots .$$Subtracting the first equation from the second gives us $$\begin{array}{rl}(k-1)S& =4+\frac{5}{k}+\frac{5}{k^2}+\frac{5}{k^3}+\cdots \\ & =4+\frac{\frac{5}{k}}{1-\frac{1}{k}}\\ & =4+\frac{5}{k-1}\\ & =\frac{4k+1}{k-1}.\end{array}$$Therefore, $$S=\frac{4k+1}{(k-1{)}^2}=\frac{13}{4}.$$Rearranging gives, $$16k+4=13(k^2-2k+1).$$Bringing all the terms on one side gives us $$13k^2-42k+9=0$$Factoring gives $$(k-3)(13k-3)=0.$$Hence, $k=3$ or $k=\frac{3}{13}$. Since we are told that $k>1$ (and more importantly, the series converges), we have that $k=\boxed{3}.$