If \(a_{0}=\frac{1}{5}\), then we have no problem. If \(a_{0}\neq\frac{1}{5}\), \(\left(\frac{2}{3}\right)^{n}\) goes to 0 when \(n\rightarrow\infty\), so \(\frac{a_{n}}{3^{n}}\) will have the same sign as \(\left(a_{0}-\frac{1}{5}\right)\left(-1\right)^{n}\) when \(n\) is large. Hence, \(a_{n}<a_{n+1}\) will not hold then contradiction