Since ∑ n = 0 ∞ ( 1 − a n ) {\displaystyle \sum _{n=0}^{\infty }(1-a_{n})} converges, lim n → ∞ ( 1 − a n ) = 0 {\displaystyle \lim _{n\rightarrow \infty }(1-a_{n})=0} , which implies that lim n → ∞ a n = 1. {\displaystyle \lim _{n\rightarrow \infty }a_{n}=1.}
Therefore, lim n → ∞ 2 n a n = ( lim n → ∞ 2 n ) ( lim n → ∞ a n ) = ∞ ⋅ 1 = ∞ . {\displaystyle \lim _{n\rightarrow \infty }2^{n}a_{n}=(\lim _{n\rightarrow \infty }2^{n})(\lim _{n\rightarrow \infty }a_{n})=\infty \cdot 1=\infty .} This implies that ∑ n = 0 ∞ 2 n a n {\displaystyle \sum _{n=0}^{\infty }2^{n}a_{n}} diverges since lim n → ∞ 2 n a n ≠ 0. {\displaystyle \lim _{n\rightarrow \infty }2^{n}a_{n}\neq 0.}
