"If n is any positive quantity shew that \frac 1{n} > \frac 1{n+1} + \frac 1{{(n+2)}^2} + \frac 3{{(n+3)}^3} + \frac {4^2}{{(n+4)}^4} + \frac {5^3}{{(n+5)}^5} + \dots Find the difference approximately when n is great. Hence shew that \frac 1{1001} + \frac 1{1002^2} + \frac 3{1003^3} + \frac {4^2}{1004^4} + \frac {5^3}{1005^5} + \dots < \frac 1{1000} by 10^{-440} nearly."