It's simplest to use the comparison test. Let a_{n}=\frac{n-1}{n\cdot 4^{n}}an=n−1n⋅4n and let b_{n}=\frac{1}{4^{n}}bn=14n. Note that 0\leq a_{n}\leq b_{n}0≤an≤bn for all integers n\geq 1n≥1 since \frac{n-1}{n}\leq 1n−1n≤1 for all integers n\geq 1n≥1.
Also note that \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}(\frac{1}{4})^{n}∞∑n=1bn=∞∑n=1(14)n converges since it is geometric with common ratio r=1/4r=14, which satisfies |r|<1|r|<1 (in fact, it converges to \frac{1/4}{1-1/4}=\frac{1}{3}141−14=13).
The Comparison Test can now be used to say that \sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{n-1}{n\cdot 4^{n}}∞∑n=1an=∞∑n=1n−1n⋅4n converges.
