$ a_{n + 1} = a_{n} - 1 + \frac {1}{a_{n} + 1}$. We see, then, that $ a_{k} = 1994 - k + \frac {1}{a_{k - 1} + 1} + \frac {1}{a_{k - 2} + 1} + ... + \frac {1}{1994 + 1}$ It remains to establish an upper bound for the "fractional part." It is plain from our expression that $ a_{k} > 1994 - k$; hence, $ \frac {1}{a_{k - 1} + 1} + \frac {1}{a_{k - 2} + 1} + ... + \frac {1}{1994 + 1} < \frac {1}{1995 - k} + \frac {1}{1995 - (k - 1)} + ... + \frac {1}{1995}$ Since these "fractional parts" are strictly increasing, it suffices to evaluate it for $ k = 998$. In this case, our upper bound is $ \frac {1}{997} + \frac {1}{998} + ... + \frac {1}{1995}$ We want to show that this is strictly less than $ 1$. There are probably several ways to do this, but I'm going to use the fact that $ \lim_{n \to \infty}\left( \frac {1}{n + 1} + ... + \frac {1}{2n}\right) = \ln 2 < 1$

Quote: I'm going to use the fact that $\lim_{n \to \infty}\left( \frac{1}{n+1}+...+\frac{1}{2n}\right) = \ln 2 < 1$ Sorry to nit-pick, but this fact isn't the important one: what's relevant is that the sum increases monotonically and doesn't get bigger than one. Its limiting behavior is irrelevant. (I know you know this, I just thought it should be said.)

Anyway, the claim that the expression is $< 1$ is a trivial one (just note that $\frac{1}{n+k}< \frac{1}{n}$ and you are done.)

t0rajir0u wrote: $ a_{n + 1} = a_{n} - 1 + \frac {1}{a_{n} + 1}$. We see, then, that $ a_{k} = 1994 - k + \frac {1}{a_{k - 1} + 1} + \frac {1}{a_{k - 2} + 1} + ... + \frac {1}{1994 + 1}$ It remains to establish an upper bound for the "fractional part." It is plain from our expression that $ a_{k} > 1994 - k$; hence, $ \frac {1}{a_{k - 1} + 1} + \frac {1}{a_{k - 2} + 1} + ... + \frac {1}{1994 + 1} < \frac {1}{1995 - k} + \frac {1}{1995 - (k - 1)} + ... + \frac {1}{1995}$ Since these "fractional parts" are strictly increasing, it suffices to evaluate it for $ k = 998$. In this case, our upper bound is $ \frac {1}{997} + \frac {1}{998} + ... + \frac {1}{1995}$ We want to show that this is strictly less than $ 1$. There are probably several ways to do this, but I'm going to use the fact that $ \lim_{n \to \infty}\left( \frac {1}{n + 1} + ... + \frac {1}{2n}\right) = \ln 2 < 1$ Nice same approach but writing this was hard, thanks for the help Also :- To prove the end, we can :- assume everything to be 1/997 then the sum would be 998/997 , but then remove 2/997 and replace it with 1/1995 and 1/1994 (not necessarily) then this sum would be larger than the previous but smaller than 1 QED