Quay lại đáp án

Ta có:A = $\frac{1}{5} + \frac{1}{13} + \frac{1}{25} + ... + \frac{1}{n^2+(n+1)^2}$A = $\frac{1}{1^2+2^2} + \frac{1}{2^2+3^2} + \frac{1}{3^2+4^2} + ... + \frac{1}{n^2+(n+1)^2}$Lại có: $n^2 + (n+1)^2 > 2n(n+1)$ ( BĐT Cô-si )=> $A < \frac{1}{2.1.2} + \frac{1}{2.2.3} + \frac{1}{2.3.4} + ... +\frac{1}{2.n.(n+1)} )$ $A < \frac{1}{2}\times ( \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + ... + \frac{1}{n.(n+1)})$ $A < \frac{1}{2}\times (1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4} + ... + \frac{1}{n} - \frac{1}{n+1})$ $A < \frac{1}{2}\times ( 1 - \frac{1}{n+1}) < \frac{1}{2}$

Ta có:A = $\frac{1}{5} + \frac{1}{13} + \frac{1}{25} + ... + \frac{1}{n^2+(n+1)^2}$A = $\frac{1}{1^2+2^2} + \frac{1}{2^2+3^2} + \frac{1}{3^2+4^2} + ... + \frac{1}{n^2+(n+1)^2}$Lại có: $n^2 + (n+1)^2 > 2n(n+1)$ ( BĐT Cô-si và n < n+1 )=> $A < \frac{1}{2.1.2} + \frac{1}{2.2.3} + \frac{1}{2.3.4} + ... +\frac{1}{2.n.(n+1)} )$ $A < \frac{1}{2}\times ( \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + ... + \frac{1}{n.(n+1)})$ $A < \frac{1}{2}\times (1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4} + ... + \frac{1}{n} - \frac{1}{n+1})$ $A < \frac{1}{2}\times ( 1 - \frac{1}{n+1}) < \frac{1}{2}$