Since ∑ n = 1 ∞ n a n − 2 n + 1 n + 1 {\displaystyle \displaystyle \sum _{n=1}^{\infty }{\frac {na_{n}-2n+1}{n+1}}} converges, we have lim n → ∞ n a n − 2 n + 1 n + 1 = 0 {\displaystyle \displaystyle \lim _{n\rightarrow \infty }{\frac {na_{n}-2n+1}{n+1}}=0} . Hence:
0 = lim n → ∞ n a n − 2 n + 1 n + 1 = lim n → ∞ ( ( a n − 2 ) ⋅ n n + 1 + 1 n + 1 ) = lim n → ∞ ( a n − 2 ) ⋅ n n + 1 + lim n → ∞ 1 n + 1 = lim n → ∞ ( a n − 2 ) ⋅ n n + 1 + 0 = lim n → ∞ ( a n − 2 ) ⋅ lim n → ∞ n n + 1 = lim n → ∞ ( a n − 2 ) ⋅ 1 = lim n → ∞ ( a n − 2 ) {\displaystyle {\begin{aligned}0&=\lim _{n\rightarrow \infty }{\frac {na_{n}-2n+1}{n+1}}\\&=\lim _{n\rightarrow \infty }\left((a_{n}-2)\cdot {\frac {n}{n+1}}+{\frac {1}{n+1}}\right)\\&=\lim _{n\rightarrow \infty }(a_{n}-2)\cdot {\frac {n}{n+1}}+\lim _{n\rightarrow \infty }{\frac {1}{n+1}}\\&=\lim _{n\rightarrow \infty }(a_{n}-2)\cdot {\frac {n}{n+1}}+0\\&=\lim _{n\rightarrow \infty }(a_{n}-2)\cdot \lim _{n\rightarrow \infty }{\frac {n}{n+1}}\\&=\lim _{n\rightarrow \infty }(a_{n}-2)\cdot 1\\&=\lim _{n\rightarrow \infty }(a_{n}-2)\\\end{aligned}}}
Therefore lim n → ∞ ( a n − 2 ) = 0 {\displaystyle \displaystyle \lim _{n\rightarrow \infty }(a_{n}-2)=0} . Or rather, lim n → ∞ a n = 2 {\displaystyle \displaystyle \lim _{n\rightarrow \infty }a_{n}=2} .
On the other hand, ln ( a n a n + 1 ) = ln a n − ln a n + 1 {\displaystyle \ln \left({\frac {a_{n}}{a_{n+1}}}\right)=\ln a_{n}-\ln a_{n+1}} . Hence,
− ln a 1 + ∑ n = 1 k ln ( a n a n + 1 ) = − ln a 1 + ∑ n = 1 k ( ln a n − ln a n + 1 ) = − ln a 1 + ( ln a 1 − ln a 2 ) + ( ln a 2 − ln a 3 ) + ⋯ + ( ln a k − ln a k + 1 ) = − ln a k + 1 {\displaystyle {\begin{aligned}-\ln a_{1}+\sum _{n=1}^{k}\ln \left({\frac {a_{n}}{a_{n+1}}}\right)&=-\ln a_{1}+\sum _{n=1}^{k}(\ln a_{n}-\ln a_{n+1})\\&=-\ln a_{1}+(\ln a_{1}-\ln a_{2})+(\ln a_{2}-\ln a_{3})+\dots +(\ln a_{k}-\ln a_{k+1})\\&=-\ln a_{k+1}\\\end{aligned}}}
Hence,
− ln a 1 + ∑ n = 1 ∞ ln ( a n a n + 1 ) = lim k → ∞ ( − ln a 1 + ∑ n = 1 k ln ( a n a n + 1 ) ) = lim k → ∞ − ln a k + 1 = − ln 2. {\displaystyle {\begin{aligned}-\ln a_{1}+\sum _{n=1}^{\infty }\ln \left({\frac {a_{n}}{a_{n+1}}}\right)&=\lim _{k\rightarrow \infty }\left(-\ln a_{1}+\sum _{n=1}^{k}\ln \left({\frac {a_{n}}{a_{n+1}}}\right)\right)\\&=\lim _{k\rightarrow \infty }-\ln a_{k+1}\\&=-\ln 2.\end{aligned}}}
converges, we have 
. Hence:
. Or rather, 
.
. Hence,
