April wrote: Sequence $ \{a_{n}\}$ is defined by $ a_{1}= 2007,\, a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}$ for $ n \ge 1.$ Prove that $ [a_{n}] =2007-n$ for $ 0 \le n \le 1004,$ where $ [x]$ denotes the largest integer no larger than $ x.$ You are wrong $ [a_{1}]=2007\not =2007-1$. I think $ a_{0}=2007$,... Let $ b_{n}=a_{n}+1$, then $ b_{n+1}=1+\frac{(b_{n}-1)^{2}}{b_{n}}=b_{n}-1+\frac{1}{b_{n}}$. Therefore $ a_{n}=a_{0}-n+\sum_{i=0}^{n-1}\frac{1}{a_{i}+1}$. Because $ a_{n}\ge a_{0}-n$ we get $ \sum_{i=0}^{n-1}\frac{1}{a_{i}+1}\le \sum_{i=0}^{n-1}\frac{1}{2008-i}<ln\frac{2008.5}{2008.5-n}<1$ if n<1280.

Rust wrote: April wrote: Sequence $ \{a_{n}\}$ is defined by $ a_{1}= 2007,\, a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}$ for $ n \ge 1.$ Prove that $ [a_{n}] =2007-n$ for $ 0 \le n \le 1004,$ where $ [x]$ denotes the largest integer no larger than $ x.$ You are wrong $ [a_{1}]=2007\not =2007-1$. I think $ a_{0}=2007$,... Let $ b_{n}=a_{n}+1$, then $ b_{n+1}=1+\frac{(b_{n}-1)^{2}}{b_{n}}=b_{n}-1+\frac{1}{b_{n}}$. Therefore $ a_{n}=a_{0}-n+\sum_{i=0}^{n-1}\frac{1}{a_{i}+1}$. Because $ a_{n}\ge a_{0}-n$ we get $ \sum_{i=0}^{n-1}\frac{1}{a_{i}+1}\le \sum_{i=0}^{n-1}\frac{1}{2008-i}<ln\frac{2008.5}{2008.5-n}<1$ if n<1280. Why $sum_{i=0}^{n-1}\frac{1}{2008-i}<ln\frac{2008.5}{2008.5-n}$